实验五
task_1.1
#include <stdio.h> #define N 5 void input(int x[], int n); void output(int x[], int n); void find_min_max(int x[], int n, int *pmin, int *pmax); int main() { int a[N]; int max, min; printf("¼Èë%d¸öÊý¾Ý:\n", N); input(a, N); printf("Êý¾ÝÊÇ: \n"); output(a, N); printf("Êý¾Ý´¦Àí...\n"); find_min_max(a, N, &min, &max); printf("Êä³ö½á¹û:\n"); printf("min = %d, max = %d\n", min, max); return 0; } void input(int x[], int n) { int i; for(i = 0; i < n; ++i) scanf("%d", &x[i]); } void output(int x[], int n) { int i; for(i = 0; i < n; ++i) printf("%d ", x[i]); printf("\n"); } void find_min_max(int x[], int n, int *pmin, int *pmax) { int i; *pmax = *pmin = x[0]; for(i=0;i<n;i++) { if(x[i]<*pmin) { *pmin = x[i]; } else if(x[i]>*pmax) { *pmax = x[i]; } } }
问题一：find_min_max功能是找出数组中的最大最小值
问题二：*pmax和*pmin指向x[0]的地址
task_1.2
#include <stdio.h> #define N 5 void input(int x[], int n); void output(int x[], int n); int *find_max(int x[], int n); int main() { int a[N]; int* pmax; printf("¼Èë%d¸öÊý¾Ý:\n", N); input(a, N); printf("Êý¾ÝÊÇ: \n"); output(a, N); printf("Êý¾Ý´¦Àí...\n"); pmax = find_max(a, N); printf("Êä³ö½á¹û:\n"); printf("max = %d\n", *pmax); return 0; } void input(int x[], int n) { int i; for(i = 0; i < n; ++i) scanf("%d", &x[i]); } void output(int x[], int n) { int i; for(i = 0; i < n; ++i) printf("%d ", x[i]); printf("\n"); } int *find_max(int x[], int n) { int max_index = 0; int i; for(i=0;i<n;i++) { if(x[i]>max_index) { max_index = i; } } return &x[max_index]; }
问题一：返回的是最大值的地址
问题二：彳亍
task_2.1
#include<stdio.h> #include<string.h> #define N 80 int main() { char s1[N] = "learning makes me happy"; char s2[N] = "learning makes me sleepy"; char tmp[N]; printf("sizeof(s1) vs. strlen(s1): \n"); printf("sizeof(s1) = %d\n", sizeof(s1)); printf("strlen(s1) = %d\n", strlen(s1)); printf("\nbefore swap: \n"); printf("s1: %s\n", s1); printf("s2: %s\n", s2); printf("\nswapping...\n"); strcpy(tmp, s1); strcpy(s1, s2); strcpy(s2, tmp); printf("\nafter swap: \n"); printf("s1: %s\n", s1); printf("s2: %s\n", s2); return 0; }
问题1：s1的大小是80；sizeof(s1)表示的是包含\0的元素个数；strlen(s1)统计的是不包含\0的元素个数
问题2：不能，s1是地址，后面的是字符串
task_2.2
#include<stdio.h> #include<string.h> #define N 80 int main() { const char *s1 = "learning makes me happy"; const char *s2 = "learning makes me sleepy"; const char *tmp; printf("sizeof(s1) vs. strlen(s1): \n"); printf("sizeof(s1) = %d\n", sizeof(s1)); printf("strlen(s1) = %d\n", strlen(s1)); printf("\nbefore swap: \n"); printf("s1: %s\n", s1); printf("s2: %s\n", s2); printf("\nswapping...\n"); tmp = s1; s1 = s2; s2 = tmp; printf("\nafter swap: \n"); printf("s1: %s\n", s1); printf("s2: %s\n", s2); return 0; }
问题1：s1存放的是第一个元素的地址；sizeof(s1)计算的是指针的字节；strlen(s1)统计的是非\0元素的个数
问题2：可以；2.1中是把字符串赋值给地址，2.2中是将第一个元素的地址赋值给s1
问题3：交换的是地址，内存没有交换
task_3
#include<stdio.h>
int main()
{
int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
int i, j;
int *ptr1;
int(*ptr2)[4];
printf("Êä³ö1: ʹÓÃÊý×éÃû¡¢Ï±êÖ±½Ó·ÃÎʶþάÊý×éÔªËØ\n");
for(i = 0; i < 2; i++)
{
for (j = 0; j < 4; j++)
printf("%d ", x[i][j]);
printf("\n");
}
printf("\nÊä³ö2: ʹÓÃÖ¸Õë±äÁ¿ptr1(Ö¸ÏòÔªËØ)¼ä½Ó·ÃÎÊ\n");
for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ptr1++, i++)
{
printf("%d", *ptr1);
if((i+1)%4==0)
printf("\n");
}
printf("\nÊä³ö3: ʹÓÃÖ¸Õë±äÁ¿ptr2(Ö¸ÏòһάÊý×é)¼ä½Ó·ÃÎÊ\n");
for (ptr2 = x; ptr2 < x + 2; ptr2++)
{
for(j=0;j<4;j++)
{
printf("%d", *(*ptr2+j));
}
printf("\n");
}
return 0;
}
*(ptr)中ptr是数组指针，指向该数组的指针；另一个是指针数组，数组内元素均为指针
task_4
#include<stdio.h> #define N 80 void replace(char *str, char old_char, char new_char); int main() { char text[N] = "Programming is difficult or not, it is a question."; printf("原始文本: \n"); printf("%s\n", text); replace(text, 'i', '*'); printf("处理后文本: \n"); printf("%s\n", text); return 0; } void replace(char *str, char old_char, char new_char) { int i; while(*str!='\0') { if(*str == old_char) *str = new_char; str++; } }
问题1：replace功能是把指定字母换成*
问题2：彳亍
task_5
#include<stdio.h> #define N 80 char *str_trunc(char *str, char x); int main() { char str[N]; char ch; while(printf("输入字符串: "),gets(str)!=NULL) { printf("输入一个字符: "); ch = getchar(); printf("截断处理...\n");
str_trunc(str, ch); printf("处理后的字符串: %s\n\n", str); getchar(); } return 0; } char *str_trunc(char *str, char x) { char *tmp = str; for(;*tmp!=x;tmp++); *tmp = '\0'; return str; }
问题：去掉gechar()后，从第二组开始没法输入字符串，起到的作用是存储输入字符后的回车
task_6
#include<stdio.h>
#include<string.h>
#define N 5
int check_id(char *str);
int check_central(char *str);
int check_last(char *str);
int main()
{
char *pid[N] = {"31010120000721656X",//T
"3301061996X0203301",//F
"53010220051126571", //F
"510104199211197977",//T
"53010220051126133Y"};//F
int i;
for (i = 0; i < N; i++)
if (check_id(pid[i]))
printf("%s\tTrue\n", pid[i]);
else
printf("%s\tFalse\n", pid[i]);
return 0;
}
int check_id(char *str)
{
int i;
if(strlen(str)==18 && check_central(str)!=0 && check_last(str)==1)
{
return 1;
}
else
{
return 0;
}
}
int check_central(char *str)
{
int i;
for(i=0;i<17;i++)
{
if(*(str+i)>='A' && *(str+i)<='Z')
{
return 0;
}
}
}
int check_last(char *str)
{
if((*(str+17)>='0' && *(str+17)<='9') || *(str+17) == 'X')
return 1;
}
task_7
#include<stdio.h> #include<string.h> #define N 80 void encoder(char *str, int n); void decoder(char *str, int n); int main() { char words[N]; int n; printf("输入英文文本:"); gets(words); printf("输入n:"); scanf("%d", &n); printf("编码后的英文文本:"); encoder(words, n); printf("%s\n", words); printf("解码后的英文文本:"); decoder(words, n); printf("%s\n", words); return 0; } void encoder(char *str, int n) { int i; for(i=0;*(str+i)!='\0';i++) { if(*(str+i) >= 'a' && *(str+i) <= 'z' - n || *(str+i) >= 'A' && *(str+i) <= 'Z' - n) { *(str+i) += n; } else if(*(str+i) > 'z' - n && *(str+i) <= 'z' || *(str+i) > 'Z' - n && *(str+i) <= 'Z') { *(str+i) = *(str+i) + n - 26; } } } void decoder(char *str, int n) { int i; for(i=0;*(str+i)!='\0';i++) { if(*(str+i) >= 'a' + n && *(str+i) <= 'z' || *(str+i) >= 'A' + n && *(str+i) <= 'Z') { *(str+i) -= n; } else if(*(str+i) >= 'a' && *(str+i) < 'a' + n || *(str+i) >= 'A' && *(str+i) < 'A' + n) { *(str+i) = *(str+i) - n + 26; } } }
task_8
#include<stdio.h> int main(int argc, char *argv[]) { int i, j; char *tmp; for(i=1;i<argc-1;i++) { for(j=1;j<argc;j++) { if(argv[i]>argv[j]) { tmp = argv[i]; argv[i] = argv[j]; argv[j] = tmp; } } } for(i=1;i<argc;i++) { printf("Hello, %s\n", argv[i]); } return 0; }
