实验5
实验一
代码:
#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 min, max; printf("录入%d个数据:\n", N); input(a, N);
#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("录入%d个数据:\n", N); input(a, N); printf("数据是: \n"); output(a, N); printf("数据处理...\n"); pmax = find_max(a, N); printf("输出结果:\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] > x[max_index]) max_index = i; return &x[max_index]; }
printf("数据是: \n"); output(a, N); printf("数据处理...\n"); find_min_max(a, N, &min, &max); printf("输出结果:\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; *pmin = *pmax = 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1:find_min_max的作用是将最大值的地址赋给pmax,最小值的地址赋给pmin,pmin和pmax都·指向x[0]
问题2:find_max的功能是返回最大值的地址;可以写成以下代码
实验二
代码:
#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:s1的大小是23;sizeof(s1)计算的是s1所占字节数;strlen(s1)统计的是s1的字符串长度;不能,定义了一个数组后不能直接赋值,可以用strcpy函数;内容已交换;
#include <stdio.h> #include <string.h> #define N 80 int main() { char *s1 = "Learning makes me happy"; char *s2 = "Learning makes me sleepy"; 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:存放的是字符串 Learning makes me happy 在内存中的地址;sizeof(s1) 计算的是字符串 Learning makes me happy 在内存中的地址名 的大小;strlen(s1) 统计的是字符串 Learning makes me happy的长度;能替换,
char *s1; s1 = "Learning makes me happy";
定义了一个字符指针并让它指向一个字符串常量,由于s1只代表地址,所以不能修改字符串常量的内容,而2.1中是定义了一个字符数组,并将字符串复制到数组中,数组中的字符串可以被修改;交换的是两个字符串的地址;没有交换;
实验三
代码:
#include <stdio.h> int main() { int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}}; int i, j; int *ptr1; // 指针变量,存放int类型数据的地址 int(*ptr2)[4]; // 指针变量,指向包含4个int元素的一维数组 printf("输出1: 使用数组名、下标直接访问二维数组元素\n"); for (i = 0; i < 2; ++i) { for (j = 0; j < 4; ++j) printf("%d ", x[i][j]); printf("\n"); } printf("\n输出2: 使用指针变量ptr1(指向元素)间接访问\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("\n输出3: 使用指针变量ptr2(指向一维数组)间接访问\n"); for (ptr2 = x; ptr2 < x + 2; ++ptr2) { for (j = 0; j < 4; ++j) printf("%d ", *(*ptr2 + j)); printf("\n"); } return 0; }
问题:ptr1存放地址,ptr2指向数组
实验四
代码:
#include <stdio.h> #define N 80 void replace(char *str, char old_char, char new_char); // 函数声明 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', '*'); // 函数调用 注意字符形参写法,单引号不能少 printf("处理后文本: \n"); printf("%s\n", text); return 0; } // 函数定义 void replace(char *str, char old_char, char new_char) { int i; while(*str) { if(*str == old_char) *str = new_char; str++; } }
问题:将文本中的i替换为*;可以
实验五:
代码:
#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); // 函数调用 printf("截断处理后的字符串: %s\n\n", str); getchar(); } return 0; } char *str_trunc(char *str,char x){ char *p=str; while(*p!='\0'){ if(*p==x){ *p='\0'; break;} p++; } }
问题:使回车不被读取
实验六
代码:
#include <stdio.h> #include <string.h> #define N 5 int check_id(char *str); // 函数声明 int main() { char *pid[N] = {"31010120000721656X", "3301061996X0203301", "53010220051126571", "510104199211197977", "53010220051126133Y"}; int i; for (i = 0; i < N; ++i) if (check_id(pid[i])) // 函数调用 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) return 0; else { for(i=0;i<=17;i++) if('0'<=str[i]<='9') break; else return 0;} if(str[17]=='x'||'0'<=str[17]<='9') return 1; }
实验七
代码:
#include <stdio.h> #define N 80 void encoder(char *str, int n); // 函数声明 void decoder(char *str, int n); // 函数声明 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); // 函数调用 printf("%s\n", words); printf("对编码后的英文文本解码: "); decoder(words, n); // 函数调用 printf("%s\n", words); return 0; } void encoder(char *str, int n) { int i=0; while(str[i]!='\0'){ if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z')) { if((str[i]>='a'&&str[i]+n<='z')||(str[i]>='A'&&str[i]+n<='Z')){ str[i]+=n; } else{ if(str[i]>='a') str[i]='a'+str[i]+n-'z'-1; else str[i]='A'+str[i]+n-'Z'-1; } } i++; } } void decoder(char *str, int n) { int i=0; while(str[i]!='\0'){ if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z')) { if(str[i]>='a'&&str[i]+n<='z'){ str[i]=((str[i]-'a'-n+26)%26)+'a';} else str[i]=((str[i]-'A'-n+26)%26)+'A'; } i++; } }
实验八
代码:
#include <stdio.h> #include<string.h> int main(int argc, char *argv[]) { int i,j; char *t; for(i=1;i<argc-1;i++){ for(j=1;j<argc-1-i;j++){ if(strcmp(argv[j],argv[j+1])>0){ t=argv[j]; argv[j]=argv[j+1]; argv[j+1]=t; } } } for(i = 1; i < argc; ++i) printf("hello, %s\n", argv[i]); return 0; }