实验5
任务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("输出结果是:\n"); printf("min=%d,max=%d", min, max); return 0; } void input(int x[], int n) { int i; for (i = 0;i < n;++i) scanf_s("%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]; }
1 找出最大最小值
2 代表每个数组元素地址
#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]; }
找最大值，并取其地址
可以
任务2
#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 ，sizeof计算的是定义数组时的长度，strlen统计的是数组的实际长度包括空格
s1是地址，不可赋值字符串
没
#include<stdio.h> #include<string.h> #define N 80 #define _CRT_SECURE_NO_WARNINGS 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; }
s1是字符串，sizeof是字符串长度，strlen计算的是字符串长度包括空格
可以，指针是变量可赋值
交换两个地址，内存没
任务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("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：使用指针变量莆田人（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; }
数组指针 行指针
四个元素的指针数组 存放地址
任务4
#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++; } }
字符串中old_char的部分改为new_char
可以
任务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("截断处理后的字符串：%s\n\n",str); getchar(); } return 0; } char *str_trunc(char *str,char x){ char *copy=str; while(*str){ if(*str!=x){ *copy=*str; str++;copy++; } else{ *str='\0'; break; } } return copy; }
读取“\n”防止影响后面的多组输入
避免回车被当作有效字符
任务6
#include<stdio.h> #include<string.h> #define N 5 int check_id(char *str); int main(){ char *pid[N]={"31010120000721656x", "3301061996x0203301", "53010220021126571", "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; } int check_id(char *str){ int j,i; if(strlen(str)!=18) return 0; for(i=0;i<18;i++){ if(i==17&&str[i]=='x'){ continue; } if(str[i]<'0'||str[i]>'9') return 0; } return 1; }
任务7
#include <stdio.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("对编码后的英文文本解码: "); decoder(words, n); printf("%s\n", words); return 0; } void encoder(char *str, int n) { while (*str) { if (*str >= 'a' && *str <= 'z') { *str = (*str - 'a' + n + 26) % 26 + 'a'; } if (*str >= 'A' && *str <= 'Z') { *str = (*str - 'A' + n + 26) % 26 + 'A'; } str++; } } void decoder(char *str, int n) { while (*str) { if (*str >= 'a' && *str <= 'z') { *str = (*str - 'a' - n + 26) % 26 + 'a'; } if (*str >= 'A' && *str <= 'Z') { *str = (*str - 'A' - n + 26) % 26 + 'A'; } str++; } }
任务8
#include<stdio.h> #include<string.h> int main(int argc,char *argv[]){ int i,k,j; char *t; for(i=1;i<argc;i++){ k=i; for(j=i+1;j<argc;j++){ if(strcmp(argv[j],argv[k])<0) k=j; } if(k!=i){ t=argv[i]; argv[i]=argv[k]; argv[k]=t; } } for(i=1;i<argc;i++){ printf("hello,%s\n",argv[i]); } return 0; }
