实验五
task1-1
#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 min, max; 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"); 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]; }
find_min_max的作用：找到输入这组数据的最大值和最小值
输入数据中的第一个数据
task1-2
#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("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]; }
返回的是最大值的所在地址
可以
task2-1
#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; }
80 s1的数组大小 s1的数组长度
可以
交换
task2-2
#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; }
地址 s1所存放的地址的大小 s1中地址所指向的字符串的长度
可以，源代码是直接定义了一个指向字符串的指针，而替换是先定义一个指针然后再让他指向字符串
交换的是地址，没有
task3
#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; }
ptr为指向含4个元素的一维整型数组的指针变量
ptr为由4个指向整型数据的指针变量组成的指针数组
task4
#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++; } }
将输入的字符串中所有i替换成*
可以
task5
#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; } // 函数定义 char *str_trunc(char *str, char x) { char *p = str; while (*p!= '\0') { if (*p == x) { *p = '\0'; return str; } p++; } return str; }
如果去掉这行代码：输入一个字符后按回车键，回车键产生的换行符会留在输入缓冲区。下一次循环执行gets(str)时，它会读取这个换行符当作输入的字符串，导致程序逻辑出错，在多组输入时后续的字符串输入就会不正常。
task6
#include <stdio.h> #include <string.h> #define N 5 int check_id(char *str); int main() { char *pid[N] = {"310102120000721656X", "3301061996X0203301", "53010220051126571", "510104199211197977", "53010220051126133Y"}; 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; } // 函数定义 int check_id(char *str) { int len = strlen(str); if (len!= 18) return 0; int sum = 0; int weight[] = {7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2}; char checksum[] = {'1', '0', 'X', '9', '8', '7', '6', '5', '4', '3', '2'}; for (int i = 0; i < 17; ++i) { if (str[i] < '0' || str[i] > '9') return 0; sum += (str[i] - '0') * weight[i]; } int index = sum % 11; if (str[17] == checksum[index]) return 1; return 0; }
task7
#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; } // 函数定义 void encoder(char *str, int n) { int i; for (i = 0; str[i]!= '\0'; i++) { if (isalpha(str[i])) { if (islower(str[i])) { str[i] = ((str[i] - 'a' + n) % 26) + 'a'; } else { str[i] = ((str[i] - 'A' + n) % 26) + 'A'; } } } } void decoder(char *str, int n) { int i; for (i = 0; str[i]!= '\0'; i++) { if (isalpha(str[i])) { if (islower(str[i])) { str[i] = ((str[i] - 'a' - n + 26) % 26) + 'a'; } else { str[i] = ((str[i] - 'A' - n + 26) % 26) + 'A'; } } } }
task8
#include <stdio.h> #include <string.h> int main(int argc, char *argv[]) { int i,j; char* temp; for(i = 1; i < argc; i++){ for(j=1;j<argc-i;j++){ if(strcmp(argv[j],argv[j+1])>0){ temp = argv[j]; argv[j] = argv[j+1]; argv[j+1] = temp; } } } for(i = 1; i < argc; ++i) printf("hello, %s\n", argv[i]); return 0; }
