c: two-dimensional array in windows 10 or Ubuntu 20.4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 | #include <stdio.h> #include <stdlib.h> #include <assert.h> /** * @brief * * @param arr * @param length * @param key */ void removenum( int arr[], int length, int key); /** * @brief * * @param arr * @param length */ void printnum( int arr[], int length); /** * @brief * * @param arr * @param length */ void printnum( int arr[], int length) { assert (arr!=NULL); for ( int i=0;i<length;i++) { printf ( "%d\n" ,arr[i]); } } /** * @brief * * @param arr * @param length * @param key */ void removenum( int arr[], int length, int key) { assert (arr!=NULL); for ( int i=0;i<length;i++) { if (key==arr[i]) { for ( int j=0;j<length-1;j++) { arr[i]=arr[j+1]; } arr[length-1]=0; i--; } } } int arr[12] = { 10,20,30,40,50,60,70,80,90,100,110,120 }; int (*parr)[12] = &arr; //==int[12]* parr=&arr; *parr指向 int arr[12]的数组。数组里面存储的是12个元素的整形数据。 int * pparr = &arr[0]; //==int* pparr[12]= &arr[0],数组元素存储的是指针 void show() { printf ( "sizeof(arr)=%ld\t sizeof(parr)=%ld\n" , sizeof (arr), sizeof (parr)); //sizeof(arr)=48 sizeof(parr)=8 //特别注意的是:指针和数组,sizeof是不一样,但是访问元素的时候,可以是相同的形式 //sizeof(parr)是指针:占用的内存为8个字节是指针固定占用的内存空间。 //sizeof(arr)是数组:占用的是元素个数总和的空间。 printf ( "*arr=%d,parr=%d,*pparr=%d,\n" ,*arr,*(*parr),*parr); printf ( "arr=%p\t parr=%p\n" ,arr,parr); //都是指向数组的首地址 //parr:数组指针指向的是一个数组,准确的说,指向的是一个数组首个元素的地址 printf ( "*parr=%p\n" ,*parr); //parr是一个指针指向的是arr数组,arr数组返回的又是数组元素的手地址,所以*parr返回的是arr数组首个元素的地址 printf ( "*(*parr)=%d\n" ,*(*parr)); //10 //*(*parr)先解引用parr得到的是第一个元素的地址,在解引用指针就能获取元素的值 printf ( "parr访问第二个元素地址=%p\n" ,*parr+2); printf ( "arr+1=%p\t parr+1=%p\n" ,arr+1,parr+1); //arr指向的是数组的首地址,+1后表示下一个地址 //parr:指针指向的是arr的整个数组,parr+1后指向的就是下一个数组首个元素的地址。 for ( int j=0;j<4;j++) { } //遍历 for ( int i=0;i<12;i++) { printf ( "第%d个元素的地址=%p\t 第%d个元素的值=%d\n" ,i,*parr+i,i,*(*parr+i)); } } void machar() { int a=10; char * dChar; dChar=( char *) malloc (10); strcpy (dChar, "cbg" ); strcat (dChar, "B" ); printf ( "dChar=%p\n" ,dChar); //*dChar='A'; //strcpy(dChar,'C'); //strcat(dChar,"B"); //dChar[0]='C'; //dChar[1]='b'; //dChar[2]='G'; char * ppf; ppf=dChar; printf ( "dChar=%s\n" ,dChar); char *str; /* 最初的内存分配 */ str = ( char *) malloc (15); strcpy (str, "runoob" ); printf ( "String = %s, Address = %u\n" , str, str); /* 重新分配内存 */ str = ( char *) realloc (str, 25); strcat (str, ".com" ); printf ( "String = %s, Address = %u\n" , str, str); free (str); free (dChar); } int main() { printf ( "geovindu\n" ); printf ( "hello world!" ); printf ( "你好,世界\n" ); machar(); int arrdu[4][5]={10,20,30,40,50,60,70,80,90,100,110,120,130,140,150,160,170,180,190,200}; int arrd[3][4]={10,20,30,40,50,60,70,80,90,100,110,120}; int (*pp)[4]=arrd; //printf("%d",(*pp)[0]); //printf("%d",(*pp)[1]); //printf("%d",*pp[0]); //printf("%d",*pp[1]); //1. printf ( "1列的首位元素\n" ); for ( int k=0;k<3;k++) { printf ( " %d " ,*pp[k]); //列的首位元素 } printf ( "\n2第一行的遍历值\n" ); for ( int k=0;k<3;k++) { printf ( " %d " ,(*pp)[k]); //第一行的遍历值 } printf ( "\n3列的首位元素\n" ); //3. for ( int k=0;k<3;k++) { printf ( " %d " ,pp[k][0]); //列的首位元素 } printf ( "\n4第一行的遍历值\n" ); //4 第一行的遍历值 for ( int k=0;k<4;k++) { printf ( " %d " ,(*pp)[k]); } printf ( "*((*pp)+1)=%d\n" ,*((*pp)+1)); //50 printf ( "*pp[1]=%d\n" ,*pp[1]); //50 printf ( "(*pp[1])+1=%d\n" ,(*pp[1])+1); //51 printf ( "*(*pp+1)=%d\n" ,*(*pp+1)); //20 printf ( "*(*arrd+1)=%d\n" ,*(*arrd+1)); //50 printf ( "*arrd[1]=%d\n" ,*arrd[1]); //50 printf ( "*arrd[3]=%d\n" ,*arrd[3]); //50 printf ( "\n6指针遍历二维数组\n" ); int *dup; dup=arrd[0]; for ( int i = 0; i < sizeof (arrd) / sizeof ( int ); i++) { //printf("%d ",&arrd[i]); //p = arr[i]; printf ( "%d " ,*dup++); } printf ( "\n7遍历二维数组\n" ); //6 遍历二维数组 for ( int j=0;j<3;j++) { for ( int k=0;k<4;k++) { printf ( " %d " ,pp[j][k]); } printf ( "\n" ); } printf ( "\n8遍历二维数组\n" ); //6 遍历二维数组 for ( int j=0;j<3;j++) { for ( int k=0;k<4;k++) { printf ( " %d " ,arrd[j][k]); } printf ( "\n" ); } printf ( "\n" ); show(); char b[]= "agbdkfjdkajfkdasjfdkla" ; printf ( "%ld" , sizeof (b)); int c= sizeof (b)/ sizeof (b[0]); printf ( "%d" ,c); for ( int i = 0; i < c; i++) { printf ( "i=%d,\t b=%c,\b 内存地址=%p\n" ,i,b[i],&b[i]); } int num[5]={1,2,3,4,5}; removenum(num,5,3); printnum(num,5); system ( "pause" ); return 0; } |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 | int arrdu[4][5]={10,20,30,40,50,60,70,80,90,100,110,120,130,140,150,160,170,180,190,200}; int (*pp)[5]=arrdu; printf ( "1列的首位元素\n" ); for ( int k=0;k<4;k++) { printf ( " %d " ,*pp[k]); //列的首位元素 } printf ( "\n2第一行的遍历值\n" ); for ( int k=0;k<4;k++) { printf ( " %d " ,(*pp)[k]); //第一行的遍历值 } printf ( "\n3列的首位元素\n" ); //3. for ( int k=0;k<4;k++) { printf ( " %d " ,pp[k][0]); //列的首位元素 } printf ( "\n4第一行的遍历值\n" ); //4 第一行的遍历值 for ( int k=0;k<4;k++) { printf ( " %d " ,(*pp)[k]); } printf ( "\n6指针遍历二维数组\n" ); int *dup; dup=arrdu[0]; for ( int i = 0; i < sizeof (arrdu) / sizeof ( int ); i++) { //printf("%d ",&arrd[i]); //p = arr[i]; printf ( "%d " ,*dup++); } printf ( "\n7遍历二维数组\n" ); //7 遍历二维数组 for ( int j=0;j<4;j++) { for ( int k=0;k<5;k++) { printf ( " %d " ,pp[j][k]); } printf ( "\n" ); } printf ( "\n8遍历二维数组\n" ); //8遍历二维数组 for ( int j=0;j<4;j++) { for ( int k=0;k<5;k++) { printf ( " %d " ,arrdu[j][k]); } printf ( "\n" ); } printf ( "\n9 指针遍历二维数组\n" ); int llen=4*5; for ( int i=0;i<20;++i) { printf ( " %d " ,*(*arrdu+i)); } printf ( "\n" ); printf ( "\n10 指针遍历二维数组\n" ); int *ddpp=*arrdu; for ( int i=0;i<20;++i) { printf ( " %d " ,*(ddpp+i)); } printf ( "\n" ); printf ( "\n" ); |
进行二维数组处理进行封装:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 | /** * ***************************************************************************** * @file twoDimensional.h * @brief 二维数组 Pointers and 2-D arrays * @author geovindu,Geovin Du,涂聚文 (geovindu@163.com) * ide: vscode c11,c17 windows 10 * @date 2023-10-30 * @copyright geovindu 站在巨人的肩膀上 Standing on the Shoulders of Giants * matrix => Points to base address of two-dimensional array. Since array decays to pointer. *(matrix) => Points to first row of two-dimensional array. *(matrix + 0) => Points to first row of two-dimensional array. *(matrix + 1) => Points to second row of two-dimensional array. **matrix => Points to matrix[0][0] *(*(matrix + 0)) => Points to matrix[0][0] *(*(matrix + 0) + 0) => Points to matrix[0][0] *(*matrix + 1) => Points to matrix[0][1] *(*(matrix + 0) + 1) => Points to matrix[0][1] *(*(matrix + 2) + 2) => Points to matrix[2][2] * ***************************************************************************** */ #ifndef TWODIMENSIONAL_H_ #define TWODIMENSIONAL_H_ #include <stddef.h> #include <stdbool.h> #define BUF_LEN 100 // Length of input buffer #define COUNT 5 // Initial number of strings /** * @brief 输入字符排序 * */ void stringInputSort(); /** * @brief * * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay( const int ** arry, int row, int col); /** * @brief * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay1( const ** arry, int row, int col); /** * @brief * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay0( int arry[10][10], int row, int col); /** * @brief OK * * @param arry 二维数组 * @param intlength 行列共长度 * @return int */ int pointDisplay2( int arry[10][10], int intlength); /** * @brief * * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay3( int ** arry, int row, int col); /** * @brief Ok * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay4( int ** arry, int row, int col); /** * @brief OK * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay5( int *** arry, int row, int col); /** * @brief ok * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay6( int ** arry, int row, int col); /** * @brief 释放所有堆内存 * @param ps * @param n * */ void freeMemoryChar( char **ps, size_t n); /** * @brief 释放所有堆内存 * @param ps * @param n * */ void freeMemoryInt( int **ps, size_t n); #endif |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 | /** * ***************************************************************************** * @file twoDimensional.c * @brief 二维数组 Pointers and 2-D arrays * @author geovindu,Geovin Du,涂聚文 (geovindu@163.com) * ide: vscode c11,c17 windows 10 * @date 2023-10-30 * @copyright geovindu 站在巨人的肩膀上 Standing on the Shoulders of Giants * ***************************************************************************** */ #include <stddef.h> #include <stdio.h> #include <stdbool.h> #include <stdlib.h> #include <string.h> #include <malloc.h> #include "include/twoDimensional.h" /** * @brief 输入字符排序 * */ void stringInputSort() { char buf[BUF_LEN]; // Input buffer size_t str_count = 0; // Current string count size_t capacity = COUNT; // Current maximum number of strings char **pS = calloc (capacity, sizeof ( char *)); // Pointers to strings char ** psTemp = NULL; // Temporary pointer to pointer to char char * pTemp = NULL; // Temporary pointer to char size_t str_len = 0; // Length of a string bool sorted = false ; // Indicated when strings are sorted printf ( "Enter strings to be sorted, one per line. Press Enter to end:\n" ); // Read in all the strings char *ptr = NULL; while ( true ) { ptr = fgets (buf, BUF_LEN, stdin); if (!ptr) // Check for read error { printf ( "Error reading string.\n" ); free (pS); pS = NULL; return 1; } if (*ptr == '\n' ) break ; // Empty line check if (str_count == capacity) { capacity += capacity/4; // Increase capacity by 25% if (!(psTemp = realloc (pS, capacity))) return 1; pS = psTemp; } str_len = strnlen(buf, BUF_LEN) + 1; //strnlen_s if (!(pS[str_count] = malloc (str_len))) return 2; strcpy_s(pS[str_count++], str_len, buf); } // Sort the strings in ascending order while (!sorted) { sorted = true ; for ( size_t i = 0 ; i < str_count - 1 ; ++i) { if ( strcmp (pS[i], pS[i + 1]) > 0) { sorted = false ; // We were out of order so... pTemp= pS[i]; // swap pointers pS[i]... pS[i] = pS[i + 1]; // and... pS[i + 1] = pTemp; // pS[i + 1] } } } // Output the sorted strings printf ( "Your input sorted in ascending sequence is:\n\n" ); for ( size_t i = 0 ; i < str_count ; ++i) { printf ( "%s" , pS[i] ); free (pS[i]); // Release memory for the word pS[i] = NULL; // Reset the pointer } free (pS); // Release the memory for pointers pS = NULL; // Reset the pointer } /** * @brief 可以 * * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay( const ** arry, int row, int col) { //在main 中直接使用可以 printf ( "\n6指针遍历二维数组\n" ); int *dup; //dup= arry[0]; //*(*(arry + 0));//*(arry + 0);// for ( int i = 0; i < row; i++) //sizeof(arry) / sizeof(int) { dup= arry[i]; for ( int j=0;j<col;j++) { printf ( "%d " ,*dup++); } printf ( "\n" ); } printf ( "\n" ); } /** * @brief * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @ return int */ int pointDisplay1( const ** arry, int row, int col) { //在main 中直接使用可以 printf ( "\n7指针遍历二维数组\n" ); int * dup; for ( int i = 0; i < row; i++) //sizeof(arry) / sizeof(int) { dup=arry[i]; //*arry;// for ( int j=0;j<col;j++) { // printf ("%d \t", *(dup+i)); //printf("\n"); //显示了第一行 printf ( "%d \t" , *(dup+j)); } //printf("%d ",*dup++); printf ( "\n" ); } printf ( "\n" ); } /** * @brief * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @ return int */ int pointDisplay0( int arry[10][10], int row, int col) { printf ( "\n14指针遍历二维数组\n" ); int *dup; dup=&arry[0][0]; for ( int i=0; i<row; i++){ for ( int j=0; j<col; j++){ printf ( "%d \t" , *(dup+i*col+j)); } printf ( "\n" ); } int (*pp)[col]=arry; printf ( "\n1列的首位元素\n" ); for ( int k=0;k<row;k++) { printf ( " %d " ,*pp[k]); //列的首位元素 } printf ( "\n" ); printf ( "\n2第一行的遍历值\n" ); for ( int k=0;k<row;k++) { printf ( " %d " ,(*pp)[k]); //第一行的遍历值 } printf ( "\n" ); } /** * @brief * * @param arry 二维数组 * @param intlength 行列共长度 row*col * @return int */ int pointDisplay2( int arry[10][10], int intlength) { printf ( "\n9 指针遍历二维数组\n" ); //int llen=4*5; for ( int i=0;i<intlength;++i) { printf ( " %d\t" ,*(*arry+i)); } printf ( "\n" ); } /** * @brief 可以 * * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay3( int ** arry, int row, int col) { //在main 中直接使用可以 printf ( "\n10 指针遍历二维数组\n" ); int *ddpp; //=*arry; for ( int i=0;i<row;++i) { ddpp=*(arry+i); for ( int j = 0; j < col; j++) { printf ( " %d " ,*(ddpp+j)); } printf ( "\n" ); } printf ( "\n" ); } /** * @brief Ok * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay4( int ** arry, int row, int col) { printf ( "\n11 指针遍历二维数组\n" ); for ( int i = 0; i < row; i++) { for ( int j = 0; j < col; j++) { printf ( "%d " , arry[i][j]); } printf ( "\n" ); } printf ( "\n" ); } /** * @brief ok * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay6( int ** arry, int row, int col) { printf ( "\n13 指针遍历二维数组\n" ); for ( int i = 0; i < row; i++) { //printf("Address of %d th array %u \n",i , *(arry + i)); for ( int j = 0; j < col; j++) { printf ( "%d " , *( *(arry + i) + j)); } printf ( "\n" ); } printf ( "\n" ); } /** * @brief OK * @param arry 二维数组 * @param row 行长度 * @param col 列长度 * @return int */ int pointDisplay5( int *** arry, int row, int col) { printf ( "\n12 指针遍历二维数组\n" ); for ( int i = 0; i <row; i++) { for ( int j = 0; j <col; j++) { printf ( "%d " , *arry[i][j]); } printf ( "\n" ); } printf ( "\n" ); } /** * @brief 释放所有堆内存 * @param ps * @param n * */ void freeMemoryChar( char **ps, size_t n) { for ( size_t i=0;i<n;n++) { free (ps[i]); ps[i]=NULL; } free (ps); ps=NULL; } /** * @brief 释放所有堆内存 * @param ps * @param n * */ void freeMemoryInt( int **ps, size_t n) { for ( size_t i=0;i<n;n++) { free (ps[i]); ps[i]=NULL; } free (ps); ps=NULL; } |
调用:
windows10:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 | int main() { printf ( "hello c world \n" ); printf ( "你好,中国\n" ); // stringInputSort(); int arrdu[5][4]={ {10,20,30,40}, {50,60,70,80}, {90,100,110,120}, {130,140,150,160}, {170,180,190,200} }; // 4 列 int dum=4; //5 行 int dun=5; for ( int i = 0; i <dun ; i++) { for ( int j = 0; j < dum; j++) { printf ( "%d " , arrdu[i][j]); } printf ( "\n" ); } printf ( "\n\n" ); for ( int i = 0; i <dun ; i++) { printf ( "Address of %d th array %u \n" ,i , *(arrdu + i)); for ( int j = 0; j <dum ; j++) { printf ( "arr[%d][%d]=%d\n" , i, j, *( *(arrdu + i) + j) ); } printf ( "\n\n" ); } int * ptr = malloc ((dum * dun) * sizeof ( int )); /* Putting 1 to 12 in the 1D array in a sequence */ for ( int i = 0; i < dun * dum; i++) ptr[i] = i + 1; //int** pe; //pe=arrdu; /**/ //分配内存 int ** pe = ( int **) malloc ( sizeof ( int )*dum); for ( int i=0; i<dun; i++) { pe[i] = ( int *) malloc ( sizeof ( int )*dum); } //初始化内存 //memset(*pe, 0, sizeof(int)*dum*dun); //2分配内存 int *** arr2 = malloc (dum * sizeof ( int **)); for ( int i = 0; i < dun; i++) arr2[i] = malloc (dun * sizeof ( int *)); // Initialising each element of the // pointer array with the address of // element present in the other array for ( int i = 0; i <dun; i++) { for ( int j = 0; j <dum ; j++) { arr2[i][j] = &arrdu[i][j]; } } printf ( "The values are\n" ); for ( int i = 0; i <dun ; i++) { for ( int j = 0; j <dum ; j++) { printf ( "%d " , *arr2[i][j]); } printf ( "\n" ); } //strcpy(pe,arrdu); for ( int i = 0; i <dun ; i++) { for ( int j = 0; j <dum ; j++) { pe[i][j]= arrdu[i][j]; //ptr[i][j]=arrdu[i][j]; //strcpy(pe[i][j],arrdu[i][j]); printf ( "%d\n" ,arrdu[i][j]); } } printf ( "PE The values are\n" ); for ( int i = 0; i < dun; i++) { for ( int j = 0; j <dum ; j++) { printf ( "%d " , pe[i][j]); } printf ( "\n" ); } pointDisplay0(arrdu,dun,dum); //ok pointDisplay4(pe,dun,dum); //ok pointDisplay5(arr2,dun,dum); //ok pointDisplay6(pe,dun,dum); //ok pointDisplay2(arrdu,dum*dun); //ok pointDisplay3(pe,dun,dum); pointDisplay(pe,dun,dum); //12 pointDisplay1(pe,dun,dum); //12 //释放资源 free (pe); free (arr2); pe=NULL; arr2=NULL; system ( "pause" ); // linux 无效 ,只win 下有效 return 0; } |
windows or Ubuntu:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 | printf ( "hello c world, \n" ); printf ( "你好,中国\n" ); //setlocale(LC_ALL,"CN"); //int arrdu[5][4]={10,20,30,40,50,60,70,80,90,100,110,120,130,140,150,160,170,180,190,200}; int arrdu[5][4]={ {10,20,30,40}, {50,60,70,80}, {90,100,110,120}, {130,140,150,160}, {170,180,190,200} }; //5 行 int dun=5; // 4 列 int dum=4; for ( int i = 0; i <dun ; i++) { for ( int j = 0; j < dum; j++) { printf ( "%d " , arrdu[i][j]); } printf ( "\n" ); } printf ( "\n\n" ); for ( int i = 0; i <dun ; i++) { printf ( "Address of %d th array %u \n" ,i , *(arrdu + i)); for ( int j = 0; j <dum ; j++) { printf ( "arr[%d][%d]=%d\n" , i, j, *( *(arrdu + i) + j) ); } printf ( "\n\n" ); } int * ptr = malloc ((dum * dun) * sizeof ( int )); /* Putting 1 to 12 in the 1D array in a sequence */ for ( int i = 0; i < dun * dum; i++) ptr[i] = i + 1; //int** pe; //pe=arrdu; /**/ //分配内存 // int** pe = (int**)malloc(sizeof(int)*dum); // for(int i=0; i<dun; i++) //{ //pe[i] = (int*)malloc(sizeof(int)*dum); //} //分配内存 //int** pe = (int**)malloc(sizeof(int)*dum); int ** pe = malloc (dum * sizeof ( int **)); // (int**)malloc(sizeof(int)*dum); for ( int i=0; i<dun; i++) { pe[i] = malloc (dun * sizeof ( int *)); // (int*)malloc(sizeof(int)*dun); //dum } //初始化内存 //memset(*pe, 0, sizeof(int)*dum*dun); //2分配内存 int *** arr2 = malloc (dum * sizeof ( int **)); for ( int i = 0; i < dun; i++) arr2[i] = malloc (dun * sizeof ( int *)); // Initialising each element of the // pointer array with the address of // element present in the other array for ( int i = 0; i <dun; i++) { for ( int j = 0; j <dum ; j++) { arr2[i][j] = &arrdu[i][j]; } } printf ( "The values are\n" ); for ( int i = 0; i <dun ; i++) { for ( int j = 0; j <dum ; j++) { printf ( "%d " , *arr2[i][j]); } printf ( "\n" ); } //strcpy(pe,arrdu); for ( int i = 0; i <dun ; i++) { for ( int j = 0; j <dum ; j++) { pe[i][j]= arrdu[i][j]; //ptr[i][j]=arrdu[i][j]; //strcpy(pe[i][j],arrdu[i][j]); printf ( "%d\n" ,arrdu[i][j]); } } printf ( "PE The values are\n" ); for ( int i = 0; i < dun; i++) { for ( int j = 0; j <dum ; j++) { printf ( "%d " , pe[i][j]); } printf ( "\n" ); } pointDisplay0(arrdu,dun,dum); //ok pointDisplay4(pe,dun,dum); //ok pointDisplay5(arr2,dun,dum); //ok pointDisplay6(pe,dun,dum); //ok pointDisplay2(arrdu,dum*dun); //ok pointDisplay3(pe,dun,dum); pointDisplay(pe,dun,dum); //12 pointDisplay1(pe,dun,dum); //12 |
修改一下内存分配,就都可以用。
1 2 3 4 5 6 7 | //分配内存 //int** pe = (int**)malloc(sizeof(int)*dum); int ** pe = malloc (dum * sizeof ( int **)); // (int**)malloc(sizeof(int)*dum); for ( int i=0; i<dun; i++) { pe[i] = malloc (dun * sizeof ( int *)); // (int*)malloc(sizeof(int)*dun); //dum } |
调用也可以封装在头文件中:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 | /** * ***************************************************************************** * @file pointersHat.c * @brief Understand pointers to your hat size - if you dare * @author geovindu,Geovin Du,涂聚文 (geovindu@163.com) * ide: vscode c11,c17 windows 10 * @date 2023-10-31 * @copyright geovindu站在巨人的肩膀上 Standing on the Shoulders of Giants * ***************************************************************************** */ #include <stdio.h> #include <stdbool.h> #include <string.h> #include <stdlib.h> #include "include/pointersHat.h" #include "include/typeGame.h" #include "include/twoDimensional.h" /** * @brief 显示二维数组指针 * */ void displayTwoPoint() { int arrdu[5][4]={ {10,20,30,40}, {50,60,70,80}, {90,100,110,120}, {130,140,150,160}, {170,180,190,200} }; // 4 列 int dum=4; //5 行 int dun=5; for ( int i = 0; i <dun ; i++) { for ( int j = 0; j < dum; j++) { printf ( "%d " , arrdu[i][j]); } printf ( "\n" ); } printf ( "\n\n" ); for ( int i = 0; i <dun ; i++) { printf ( "Address of %d th array %u \n" ,i , *(arrdu + i)); for ( int j = 0; j <dum ; j++) { printf ( "arr[%d][%d]=%d\n" , i, j, *( *(arrdu + i) + j) ); } printf ( "\n\n" ); } int * ptr = malloc ((dum * dun) * sizeof ( int )); /* Putting 1 to 12 in the 1D array in a sequence */ for ( int i = 0; i < dun * dum; i++) ptr[i] = i + 1; //int** pe; //pe=arrdu; /**/ //分配内存 /** * * window int** pe = (int**)malloc(sizeof(int)*dum); for(int i=0; i<dun; i++) { pe[i] = (int*)malloc(sizeof(int)*dum); } */ //windows Ubuntu 都可以用 int ** pe = malloc (dum * sizeof ( int **)); for ( int i=0; i<dun; i++) { pe[i] = malloc (dun * sizeof ( int *)); } //初始化内存 //memset(*pe, 0, sizeof(int)*dum*dun); //2分配内存 int *** arr2 = malloc (dum * sizeof ( int **)); for ( int i = 0; i < dun; i++) arr2[i] = malloc (dun * sizeof ( int *)); // Initialising each element of the // pointer array with the address of // element present in the other array //赋值 for ( int i = 0; i <dun; i++) { for ( int j = 0; j <dum ; j++) { arr2[i][j] = &arrdu[i][j]; } } printf ( "The values are\n" ); for ( int i = 0; i <dun ; i++) { for ( int j = 0; j <dum ; j++) { printf ( "%d " , *arr2[i][j]); } printf ( "\n" ); } //strcpy(pe,arrdu); //赋值 for ( int i = 0; i <dun ; i++) { for ( int j = 0; j <dum ; j++) { pe[i][j]= arrdu[i][j]; //ptr[i][j]=arrdu[i][j]; //strcpy(pe[i][j],arrdu[i][j]); printf ( "%d\n" ,arrdu[i][j]); } } printf ( "PE The values are\n" ); for ( int i = 0; i < dun; i++) { for ( int j = 0; j <dum ; j++) { printf ( "%d " , pe[i][j]); } printf ( "\n" ); } pointDisplay0(arrdu,dun,dum); //ok pointDisplay4(pe,dun,dum); //ok pointDisplay5(arr2,dun,dum); //ok pointDisplay6(pe,dun,dum); //ok pointDisplay2(arrdu,dum*dun); //ok pointDisplay3(pe,dun,dum); //ok pointDisplay(pe,dun,dum); //ok pointDisplay1(pe,dun,dum); //ok //释放资源 free (pe); free (arr2); pe=NULL; arr2=NULL; } |
输出:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 | /** * ***************************************************************************** * @file geovindu.h * @brief * * @author geovindu,Geovin Du,涂聚文 (geovindu@163.com) * ide: vscode c11,c17 windows 10 * @date 2023-11-01 * @copyright geovindu 站在巨人的肩膀上 Standing on the Shoulders of Giants * ***************************************************************************** */ #ifndef GEOVINDU_H_ #define GEOVINDU_H_ #include <stdio.h> #include <stdbool.h> #include <ctype.h> #include <string.h> #include "CheckTieck.h" #include "TakeNumber.h" #include "Dustring.h" #include "SortAlgorithm.h" #include "KruskalAlgorithm.h" #include "FordFulkersonAlgorithm.h" #include "PrimsAlgorithm.h" #include "HuffmanCoding.h" #include "RecursionFunc.h" #include "duSortType.h" /** * @brief 排队叫号 * */ void QueueDisplay(); /** * @brief 作业 * */ void HomeWork(); /** * @brief 其他 * */ void other(); /** * @brief 1.冒泡排序 * */ void displayBubbleSort(); /** * @brief 2选择排序 * */ void displaySelectionSort(); /** * @brief 3插入排序 * */ void displayInsertionSort(); /** * @brief 4快速排序 * */ void displayQuickSort(); /** * @brief 5合并排序 * */ void displayMergeSort(); /** * @brief 6计数排序 * */ void displayCountingSort(); /** * @brief 7基数排序 * */ void displayRadixsort(); /** * @brief 8桶排序 * */ void displayBucketSort(); /** * @brief 9堆排序 * */ void displayHeapSort(); /** * @brief 10.希尔排序 * */ void displayShellSort(); /** * @brief 11.顺序查找 * */ void displayLinearSearch(); /** * @brief 12.二分搜索 * */ void displayBinarySearch(); /** * @brief 13.kruskal算法,请输入三个数字一行,输完一个数字时,按一个空格分开 * */ void displayKruskalAlgo(); /** * @brief Ford - Fulkerson algorith * */ void displayFordFulkerson(); /** * @brief Dijkstra's Algorithm 迪杰斯特拉算法 最短路径算法 * */ void displayDijkstra(); /** * @brief Prim's Algorithm * */ void displayPrims(); /** * @brief 17.霍夫曼编码 Char | Huffman code * */ void displayHuffmanarr(); #endif |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 | /** * ***************************************************************************** * @file geovindu.c * @brief * @author geovindu,Geovin Du,涂聚文 (geovindu@163.com) * ide: vscode c11,c17 windows 10 * @date 2023-11-01 * @copyright geovindu 站在巨人的肩膀上 Standing on the Shoulders of Giants * ***************************************************************************** */ #include "include/geovindu.h" /** * @brief 排队叫号 * */ void QueueDisplay() { QueueCalling *queue1; //int iii,nnn; char select= '1' ; //int num=1;//顾客序号 num=0; //叫号编号 queue1=QueueInit(); //初始化队列 if (queue1==NULL) { printf ( "创建队列时出错!\n" ); getch(); return 0; } do { if (select== '1' ||select== '2' ) //不定这条件,在Ubuntu上此列表会显示两次 { printf ( "\n请选择具体操作:\n" ); printf ( "1.新到顾客\n" ); printf ( "2.下一个顾客\n" ); printf ( "0.退出\n" ) ; fflush (stdin); } select= getchar (); //windows getch() Ubuntu: getchar() switch (select) { case '1' : add(queue1); printf ( "\n现在共有%d位顾客在等候!\n" ,QueueLen(queue1)); break ; case '2' : next(queue1); printf ( "\n现在共有%d位顾客在等候!\n" ,QueueLen(queue1)); break ; case '0' : break ; } } while (select!= '0' ); QueueFree(queue1); //释放队列 //getch(); getchar (); //内存分配函数 malloc() 分配并初始化函数 calloc() 重新分配内存函数 realloc 释放内存free() int *buf1,* buf2, * buf3; buf1=( int *) malloc (100* sizeof ( int )); buf2=( int *) calloc (100, sizeof ( int )); buf3=( int *) realloc (buf2,500* sizeof ( int )); free (buf1); free (buf3); } /** * @brief 作业 * */ void HomeWork() { char *dustr = "GXZXLeaag%^*** 1092367145 &*@654123HUYqianrushi" ; //这个不可以用替换函数,需要索引长度大于自身的宽度 char *substr = "109236714533" ; char *substr2 = "654123" ; char geovindu[100] = "GXZXLeaag%^*** 1092367145 &*@654123HUYqianrushi" ; //必须索引值大,否则不可以替换 char *newdig[10]; char *digg; char *sdu; char *newdig2[6]; char newstrd[10]={0}; //初始化赋值 char *ddu[10]={ '\0' };; //初始化赋值 int digitsdu[10] = {0}; // 存储数字出现的次数 for ( int i = 0; substr[i] != '\0' ; i++) { if ( isdigit (substr[i])) { digitsdu[substr[i] - '0' ]++; } } for ( int i = 9; i >= 0; i--) { for ( int j = 0; j < digitsdu[i]; j++) { char digitChar = i + '0' ; strncat (newstrd, &digitChar, 1); //printf("%c",digitChar); strncat (ddu, &digitChar, 1); } } char newstrd2[6]={0}; //初始化赋值 char *ddu2[6]={ '\0' }; //初始化赋值 int digitsdu2[10] = {0}; // 存储数字出现的次数 for ( int i = 0; substr2[i] != '\0' ; i++) { if ( isdigit (substr2[i])) { digitsdu2[substr2[i] - '0' ]++; } } for ( int i = 9; i >= 0; i--) { for ( int j = 0; j < digitsdu2[i]; j++) { char digitChar = i + '0' ; strncat (newstrd2, &digitChar, 1); //*(newchar+j)=digitChar; //*(newchar+i)=digitChar; //printf("%c",digitChar); //ddu2[i]=digitChar; strncat (ddu2, &digitChar, 1); } } printf ( "\n1: %s\n" ,newstrd); printf ( "2: %s\n" ,ddu); printf ( "3: %s\n" ,newstrd2); printf ( "4: %s\n" ,ddu2); char *newddd=newstrd; char *fff; char *kk; char dustr11[10] = "1092367145" ; char dustr22[6] = "654123" ; digg=duStrCmpSortDesc(dustr11,newdig,10); kk=duStrCmpSortDesc(dustr22,newdig2,6); char * Olddustr11 = "1092367145" ; char * Olddustr22 = "654123" ; fff=duReplace(geovindu,Olddustr11,ddu); //ddu printf ( "fff1:%s\n" ,fff); fff=duReplace(geovindu,Olddustr22,ddu2); printf ( "fff2:%s\n" ,fff); printf ( "\nnew:%s\n" ,digg); printf ( "newdig:%s\n" ,newdig); printf ( "newdig2:%s\n" ,newdig2); printf ( "kk:%s\n" ,kk); printf ( "fff3:%s\n" ,fff); /* //查找索引 char *dus = strstr (dustr,substr); // char *dus2= strstr (dustr,substr2); // if (dus==NULL) printf ( "can't find %s in %s\n" ,substr,dustr); else printf ( "%s include %s;show the string from start found address:%s\n" , dustr,substr,dus); //起始索引 int index=dus - dustr; int index2=dus2-dustr; //结束索引 int endindex=index+ strlen (substr); int endindex2=index2+ strlen (substr2); printf ( "1092367145 start index:%d,end:%d.\n" ,index,endindex); printf ( "654123 start index:%d,end:%d.\n" ,index2,endindex2); */ //char ch[100] = "GXZXLeaag%^*** 1092367145 &*@654123HUYqianrushi"; char duresult[100]; // 存储处理后的结果 char gx[]= "GXZX" ; char *dup; char *gxdup; char newstr[100]; char *des; dup=fff; //fff char gxz[]= "高训中心" ; //strcat(newstr, "高训中心"); strncat (newstr,&gxz,8); //printf("%s/n",gxdup); int gxlen= strlen ( "高训中心" ); //*(dup+14)='d'; int l= sizeof (geovindu)/ sizeof (geovindu[0]); //倒序显示 //for(int i=l;i>=0;i--) //{ // printf("%c\n",dup[i]); // // } //顺序显示 for ( int i=4;i<l;i++) { if (dup[i]>= 'A' && dup[i]<= 'Z' ) { printf ( "%c\n" ,dup[i]); char sd=dup[i]; strncat (newstr,&sd,1); } else if (dup[i]>= 'a' && dup[i]<= 'z' ) { printf ( "%c\n" , toupper (dup[i])); *(dup+i)= toupper (dup[i]); char sx= toupper (dup[i]); strncat (newstr,&sx,1); } else if ( isdigit (dup[i])) { printf ( "數字:%c\n" ,dup[i]); char sszi=dup[i]; strncat (newstr,&sszi,1); } else { //des=newstr; //des=dup[i]; //strcat(des[1],dup[i]); //strcpy(des, dup[i]); //memset(des, '\0', sizeof(des)); //strcat(&des,&dup[i]); //*(des+i) = '*'; //*(des+i+1)=dup[i]; //printf("%c\n",dup[i]); char qita=dup[i]; strncat (newstr, &qita, 1); strncat (newstr, &qita, 1); } // } //gxdup=duReplace(ch,"GXZX","高训中心"); //printf("gx=%s\n",gxdup); printf ( "处理后的字符串:= %s\n" ,newstr); /* // (1) 将GXZX前四个字符串中的大写字母转换成“高训中心” for ( int i = 0; i < 4; i++) { if ( isupper (ch[i])) { strcat (duresult, "高训中心" ); } else { // (2) 将字符串中其余的小写字母转换成大写字母 char uppercaseChar = toupper (ch[i]); strncat (duresult, &uppercaseChar, 1); } } // (3) 数字降序排序 int digits[10] = {0}; // 存储数字出现的次数 for ( int i = 0; ch[i] != '\0' ; i++) { if ( isdigit (ch[i])) { digits[ch[i] - '0' ]++; } } for ( int i = 9; i >= 0; i--) { for ( int j = 0; j < digits[i]; j++) { char digitChar = i + '0' ; strncat (duresult, &digitChar, 1); } } // (4) 特殊符号加倍输出 for ( int i = 4; ch[i] != '\0' ; i++) { if (! isalpha (ch[i]) && ! isdigit (ch[i])) { strncat (duresult, &ch[i], 1); strncat (duresult, &ch[i], 1); } } printf ( "处理后的字符串:%s\n" , duresult); */ } /** * @brief 其他 * */ void other() { /* textbackgroud(0); clrscr(); for(int i=1;i<8;i++) { window(10+i*5,5+i,30+i*5,15+i); textbacktgroud(i); clrscr(); } gettch();*/ /* //char a[10]="1092367145"; //变量名重复赋值,这个编译出问题 char bbdu[10]="1092367145"; char* c[10]={'\0'}; qsort(a,strlen(a),sizeof(a[0]),cmp); for(int i=0;i<=10;i++) { char du=a[i]; printf("%c ",du); strncat(c, &du, 1); //c[i]=du; } printf("\nc=:%s length=:%d\n",c,strlen(c)); char* charqs[10]={'\0'};//NULL; // char* reqs; printf("\n 原串: %s",bbdu); reqs=duStrCmpSortDesc(bbdu,charqs,10); printf("\nchar:%s lenght=:%d\n",charqs,strlen(charqs)); printf("\nchar:%s\n",reqs); */ /* //分割字符串 https://githubmota.github.io/2017/12/29/2017-12-29-Linux-C-Split/ //如何找到数字字符串的首位数和末位数索引 char ssss[] = "GXZXLeaag%^*** 1092367145 &*@654123HUYqianrushi"; char delim[] = " ,!"; char *token; for(token = strtok(ssss, delim); token != NULL; token = strtok(NULL, delim)) { printf(token); printf("\n\f"); } printf("\n"); */ //先把数字排序处理了 //再处理其他的 //最后替换其他的 //printf("%s",dup); //*(p+10)="G"; char str1[14] = "涂聚文" ; char str2[14] = "google" ; char str3[14]; int len ; /* 复制 str1 到 str3 */ strcpy (str3, str1); printf ( "strcpy( str3, str1) : %s\n" , str3 ); /* 连接 str1 和 str2 */ strcat ( str1, str2); printf ( "strcat( str1, str2): %s\n" , str1 ); /* 连接后,str1 的总长度 */ len = strlen (str1); printf ( "strlen(str1) : %d\n" , len ); printf ( "文件名:%s" __FILE__); printf ( "\n当前行号:%d" ,__LINE__); printf ( "\n日期:%s" ,__DATE__); printf ( "\n时间:%s\n" ,__TIME__); int res=RecursionFun(20); printf ( "res=%d" ,res); printf ( "\n" ); } /** * @brief 1.冒泡排序 * */ void displayBubbleSort() { int i; int *p; char str[20]; //1.冒泡排序 int data[12]={60,50,39,27,12,8,45,63,20,2,10,88}; /* 原始数据 */ int lensize= sizeof (data) / sizeof (data [0]); //sizeof(data); p=BubbleSort(data,lensize); itoa(lensize, str, 10); printf ( "\n1共長度是 %d " ,lensize); printf ( "\n1冒泡排序的结果为:" ); for (i=0;i<lensize;i++) printf ( "%3d" ,p[i]); printf ( "\n" ); } /** * @brief 2选择排序 * */ void displaySelectionSort() { int arr[] = { 64, 25, 12, 22, 11,88,28,100 }; int n = sizeof (arr) / sizeof (arr[0]); SelectionSort(arr, n); int ii; printf ( "2选择排序结果为:" ); for (ii = 0; ii < n; ii++) printf ( "%d " , arr[ii]); printf ( "\n" ); } /** * @brief 3插入排序 * */ void displayInsertionSort() { int ii; int inarr[] = {25, 23, 28, 16, 18,100,8,99}; // calculating the size of array int size = sizeof (inarr) / sizeof (inarr[0]); printf ( "3插入排序结果为:" ); InsertionSort(inarr, size); for (ii = 0; ii < size; ii++) printf ( "%d " , inarr[ii]); printf ( "\n" ); } /** * @brief 4快速排序 * */ void displayQuickSort() { int size; // defining and initializing an array int qsarr[] = {100,25, 23, 28, 16, 18,8,99,3,20}; printf ( "4快速排序结果为:" ); // calculating the size of array size = sizeof (qsarr) / sizeof (qsarr[0]); QuickSort(qsarr, 0, size - 1); for ( int i = 0; i < size; i++) printf ( "%d " , qsarr[i]); printf ( "\n" ); } /** * @brief 5合并排序 * */ void displayMergeSort() { printf ( "5合并排序结果为:" ); int mearr[] = { 12, 11, 23, 55, 6, 57,3,100,9 }; int arr_size = sizeof (mearr) / sizeof (mearr[0]); MergeSort(mearr, 0, arr_size - 1); for ( int i = 0; i < arr_size; i++) printf ( "%d " , mearr[i]); printf ( "\n" ); } /** * @brief 6计数排序 * */ void displayCountingSort() { printf ( "6计数排序结果为:" ); int carray[] = {4, 2, 2, 8, 3, 3, 1}; int cn = sizeof (carray) / sizeof (carray[0]); CountingSort(carray, cn); for ( int i = 0; i < cn; i++) printf ( "%d " , carray[i]); printf ( "\n" ); } /** * @brief 7基数排序 * */ void displayRadixsort() { printf ( "7基数排序结果为:" ); int rarray[] = {121, 432, 564, 23, 1, 45, 788}; int rn = sizeof (rarray) / sizeof (rarray[0]); Radixsort(rarray, rn); for ( int i = 0; i < rn; i++) printf ( "%d " , rarray[i]); printf ( "\n" ); } /** * @brief 8桶排序 * */ void displayBucketSort() { printf ( "8桶排序结果为:" ); int barray[] = {42, 32, 33, 5,52, 37,100, 47, 51}; BucketSort(barray); int bn = sizeof (barray) / sizeof (barray[0]); for ( int i = 0; i < bn; i++) printf ( "%d " , barray[i]); printf ( "\n" ); } /** * @brief 9堆排序 * */ void displayHeapSort() { printf ( "9堆排序结果为:" ); int harr[] = {1, 12, 9, 5, 6, 10}; int hn = sizeof (harr) / sizeof (harr[0]); HeapSort(harr, hn); for ( int i = 0; i < hn; i++) printf ( "%d " , harr[i]); printf ( "\n" ); } /** * @brief 10.希尔排序 * */ void displayShellSort() { printf ( "10.希尔排序结果为:" ); int sdata[] = {9, 8, 3, 7, 25, 6, 4, 11,38}; int ssize = sizeof (sdata) / sizeof (sdata[0]); ShellSort(sdata, ssize); for ( int i = 0; i < ssize; i++) printf ( "%d " , sdata[i]); printf ( "\n" ); } /** * @brief 11.顺序查找 * */ void displayLinearSearch() { printf ( "11.顺序查找结果为:" ); int lsdata[] = {9, 8, 3, 7, 25, 6, 4, 11,38}; int key=25; //要查找的数字 int lsize = sizeof (lsdata) / sizeof (lsdata[0]); int result = LinearSearch(lsdata, lsize,key); (result == -1) ? printf ( "\nElement not found" ) : printf ( "\nElement found at index(数组中的索引号是:): %d\n" , result); } /** * @brief 12.二分搜索 * */ void displayBinarySearch() { printf ( "\n12.二分搜索结果为:\n" ); int bsarray[] = {3, 4, 5, 6, 7, 8, 9}; int bsize = sizeof (bsarray) / sizeof (bsarray[0]); int xkey = 8; int bresult = BinarySearch(bsarray, xkey, 0, bsize - 1); if (bresult == -1) printf ( "Not found" ); else printf ( "Element is found at index(数组中的索引号是:) %d\n" , bresult); } /** * @brief 13.kruskal算法,请输入三个数字一行,输完一个数字时,按一个空格分开 * */ void displayKruskalAlgo() { printf ( "\n13.kruskal算法,请输入三个数字一行,输完一个数字时,按一个空格分开:\n" ); /* int ki; struct edge edges[N], minTree[P - 1]; for (ki = 0; ki < N; ki++) { scanf("%d %d %d", &edges[ki].initial, &edges[ki].end, &edges[ki].weight);//每行输入3个数字 输入一个数字时,按空格键 } KruskalMinTree(edges, minTree); */ //n = 6; int KruskalGraph[MAX][MAX]; KruskalGraph[0][0] = 0; KruskalGraph[0][1] = 4; KruskalGraph[0][2] = 4; KruskalGraph[0][3] = 0; KruskalGraph[0][4] = 0; KruskalGraph[0][5] = 0; KruskalGraph[0][6] = 0; KruskalGraph[1][0] = 4; KruskalGraph[1][1] = 0; KruskalGraph[1][2] = 2; KruskalGraph[1][3] = 0; KruskalGraph[1][4] = 0; KruskalGraph[1][5] = 0; KruskalGraph[1][6] = 0; KruskalGraph[2][0] = 4; KruskalGraph[2][1] = 2; KruskalGraph[2][2] = 0; KruskalGraph[2][3] = 3; KruskalGraph[2][4] = 4; KruskalGraph[2][5] = 0; KruskalGraph[2][6] = 0; KruskalGraph[3][0] = 0; KruskalGraph[3][1] = 0; KruskalGraph[3][2] = 3; KruskalGraph[3][3] = 0; KruskalGraph[3][4] = 3; KruskalGraph[3][5] = 0; KruskalGraph[3][6] = 0; KruskalGraph[4][0] = 0; KruskalGraph[4][1] = 0; KruskalGraph[4][2] = 4; KruskalGraph[4][3] = 3; KruskalGraph[4][4] = 0; KruskalGraph[4][5] = 0; KruskalGraph[4][6] = 0; KruskalGraph[5][0] = 0; KruskalGraph[5][1] = 0; KruskalGraph[5][2] = 2; KruskalGraph[5][3] = 0; KruskalGraph[5][4] = 3; KruskalGraph[5][5] = 0; KruskalGraph[5][6] = 0; edge_list spanlist; edge_list elist; KruskalAlgo(KruskalGraph,spanlist,spanlist); } /** * @brief Ford - Fulkerson algorith * */ void displayFordFulkerson() { printf ( "\n14 Ford - Fulkerson algorith \n" ); int n=10; int capacity[10][10]; for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { capacity[i][j] = 0; } } capacity[0][1] = 8; capacity[0][4] = 3; capacity[1][2] = 9; capacity[2][4] = 7; capacity[2][5] = 2; capacity[3][5] = 5; capacity[4][2] = 7; capacity[4][3] = 4; int s = 0, t =5; printf ( "\nMax Flow: %d\n" , FordFulkerson(s, t,capacity)); } /** * @brief Dijkstra's Algorithm 迪杰斯特拉算法 最短路径算法 * */ void displayDijkstra() { int Graph[MAX][MAX], j, u; int n = 7; Graph[0][0] = 0; Graph[0][1] = 0; Graph[0][2] = 1; Graph[0][3] = 2; Graph[0][4] = 0; Graph[0][5] = 0; Graph[0][6] = 0; Graph[1][0] = 0; Graph[1][1] = 0; Graph[1][2] = 2; Graph[1][3] = 0; Graph[1][4] = 0; Graph[1][5] = 3; Graph[1][6] = 0; Graph[2][0] = 1; Graph[2][1] = 2; Graph[2][2] = 0; Graph[2][3] = 1; Graph[2][4] = 3; Graph[2][5] = 0; Graph[2][6] = 0; Graph[3][0] = 2; Graph[3][1] = 0; Graph[3][2] = 1; Graph[3][3] = 0; Graph[3][4] = 0; Graph[3][5] = 0; Graph[3][6] = 1; Graph[4][0] = 0; Graph[4][1] = 0; Graph[4][2] = 3; Graph[4][3] = 0; Graph[4][4] = 0; Graph[4][5] = 2; Graph[4][6] = 0; Graph[5][0] = 0; Graph[5][1] = 3; Graph[5][2] = 0; Graph[5][3] = 0; Graph[5][4] = 2; Graph[5][5] = 0; Graph[5][6] = 1; Graph[6][0] = 0; Graph[6][1] = 0; Graph[6][2] = 0; Graph[6][3] = 1; Graph[6][4] = 0; Graph[6][5] = 1; Graph[6][6] = 0; u = 0; Dijkstra(Graph, n, u); } /** * @brief Prim's Algorithm * */ void displayPrims() { int G[V][V] = { {0, 9, 75, 0, 0}, {9, 0, 95, 19, 42}, {75, 95, 0, 51, 66}, {0, 19, 51, 0, 31}, {0, 42, 66, 31, 0}}; Prims(G); } /** * @brief 17.霍夫曼编码 Char | Huffman code * */ void displayHuffmanarr() { int arr[] = { 64, 25, 12, 22, 11,88,28,100 }; char Huffmanarr[] = { 'A' , 'B' , 'C' , 'D' }; int freq[] = {5, 1, 6, 3}; int Huffmansize = sizeof (arr) / sizeof (arr[0]); printf ( " 17.霍夫曼编码 Char | Huffman code " ); printf ( "\n--------------------\n" ); HuffmanCodes(Huffmanarr, freq, Huffmansize); } |
调用:、
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 | /* * @Author: 涂聚文 geovindu,Geovin Du * @Date: 2023-09-11 14:07:29 * @LastEditors: * @LastEditTime: 2023-09-20 14:35:49 * @FilePath: \testcpp\helloword.c * @Description: */ /*****************************************************************/ /** * \file helloworld.C * \brief 业务操作方法 * \IDE: VSCODE c11 安装插件“Doxygen Documentation Generator”,用来生成注释。 安装插件”C/C++ Snippets”,用来生成文件头、代码块分割线等。KoroFileHeader C/C++ Snippets插件设置 https://devblogs.microsoft.com/cppblog/c11-and-c17-standard-support-arriving-in-msvc/ 堆区Heap Area 栈区 Stack Area windows * \author geovindu,Geovin Du 站在巨人的肩膀上 Standing on the Shoulders of Giants * \date 2023-09-19 * \copyright * \namespace * ***********************************************************************/ #include<string.h> #include<stdio.h> #include<stdlib.h> #include<ctype.h> #include<string.h> #include<malloc.h> #include<math.h> #include<dir.h> #include<dos.h> #include<process.h> #include<dos.h> #include<bitsmsg.h> #include<time.h> #include<dos.h> #include<conio.h> #include<malloc.h> //#include "include/SortAlgorithm.h" //#include "include/KruskalAlgorithm.h" //#include "include/FordFulkersonAlgorithm.h" //#include "include/PrimsAlgorithm.h" //#include "include/HuffmanCoding.h" //#include "include/RecursionFunc.h" //#include "include/duSortType.h" // //#include "include/CheckTieck.h" //#include "include/TakeNumber.h" //#include "include/twoDimensional.h" #include "include/dynamicPrime.h" #include "include/pointersHat.h" #include "include/typeGame.h" #include "include/geovindu.h" //以文件夹的头文件能“分层”显示,易于维护和管理代码文件 int main( void ) { printf ( "hello c world \n" ); printf ( "你好,中国\n" ); //注释的函数,可以消除注释就可以执行测试效果 //findWord(); //displayGame(); // stringInputSort(); //displayPrime(); //displayHat(); //二维数组指针 //displayTwoPoint(); //排除等候 // QueueDisplay(); //作业 字符串替换 HomeWork(); // other(); //1.冒泡排序 displayBubbleSort(); //2选择排序 displaySelectionSort(); //3插入排序 displayInsertionSort(); //4快速排序 displayQuickSort(); //5 合并排序 displayMergeSort(); //6 计数排序 displayCountingSort(); //7. 基数排序 displayRadixsort(); //8 Bucket Sort 桶排序 displayBucketSort(); //9堆排序 displayHeapSort(); //10.希尔排序 displayShellSort(); //11 顺序查找(Linear/Sequential Search),也称为线性查找 displayLinearSearch(); //12 Binary Search 二分搜索 displayBinarySearch(); //13 kruskal算法 displayKruskalAlgo(); //14 Ford - Fulkerson algorith /**/ displayFordFulkerson(); //15 Dijkstra's Algorithm 迪杰斯特拉算法 最短路径算法 displayDijkstra(); //16 Prim's Algorithm displayPrims(); //17. Huffman Coding displayHuffmanarr(); system ( "pause" ); // linux 无效 ,只win 下有效 return 0; } |
哲学管理(学)人生, 文学艺术生活, 自动(计算机学)物理(学)工作, 生物(学)化学逆境, 历史(学)测绘(学)时间, 经济(学)数学金钱(理财), 心理(学)医学情绪, 诗词美容情感, 美学建筑(学)家园, 解构建构(分析)整合学习, 智商情商(IQ、EQ)运筹(学)生存.---Geovin Du(涂聚文)
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