实验5

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];
}

1找到数组x中数据的最大值和最小值赋值给对应的指针变量
2指向x[0]的地址

#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];
 }

1找到数据中的最大值，返回最大值元素的指针
2可以

task2_1.c

#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;
}

问题一：大小80字节，数组在内存占的字节数，字符串的长度

问题二：不能，数组名是指针，不能进行赋值运算

问题三：进行交换

#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;
}

task3

#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;
}

task4.c

#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) {
        if(*str == old_char)
            *str = new_char;
        str++;
    }
}

 问题1：replace的作用是将字符串中的指定字符替换成新的字符。

问题2：可以写成*str！=‘0’

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;
}
char* str_trunc(char* str, char x) {
    int n = 0;
    while (*str != 0) {
        if (*str == x)
            *str = '\0';
        else {
            str++;
            n++;
        }
    }
    return str - n;
}

task6

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define N 5

int check_id(char *str);

int main() {
    char *pid[N] = {
        "31010120000721656X",
        "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;
}

int check_id(char *str) {
    int i;

    if(strlen(str) != 18)
        return 0;

    for(i = 0; i < 17; i++) {
        if(!isdigit(str[i]))
            return 0;
    }

    if((isdigit(str[17])) || (str[17] == 'X'))
        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
功能：对str指向的字符串进行编码处理
编码规则：
对于a~z或A~Z之间的字母字符，用其后第n个字符替换; 其它非字母字符，保持不变
*/
void encoder(char* str, int n) {
    while (*str != '\0') {
        if (*str <= 'z' && *str >= 'a') {
            if (*str + n <= 'z') {
                *str += n;
            }
            else {
                *str = *str + n - 'z' - 1 + 'a';
            }
        }
        if (*str <= 'Z' && *str >= 'A') {
            if (*str + n <= 'Z') {
                *str += n;
            }
            else {
                *str = *str + n - 'Z' - 1 + 'A';
            }
        }
        str++;
    }

}

/*函数定义
功能：对str指向的字符串进行解码处理
解码规则：
对于a~z或A~Z之间的字母字符，用其前面第n个字符替换; 其它非字母字符，保持不变
*/
void decoder(char* str, int n) {
    while (*str != '\0') {
        if (*str <= 'z' && *str >= 'a') {
            if (*str - n >= 'a') {
                *str -= n;
            }
            else {
                *str = 'z' - ('a' - *str + n) + 1;
            }
        }
        if (*str <= 'Z' && *str >= 'A') {
            if (*str - n >= 'A') {
                *str -= n;
            }
            else {
                *str = 'Z' - ('A' - *str + n) + 1;
            }
        }
        str++;
    }
}

task.8

#include <stdio.h>
#include<string.h>
void sort(int n,char *s[]);

int main(int argc, char *argv[]) {
    int i;

    sort(argc-1,argv+1);


    for(i = 1; i < argc; ++i)
        printf("hello, %s\n", argv[i]);

    return 0;
}

void sort(int n,char *s[]){
    int i,j;
    char *t;

    for(i=0;i<n-1;i++){
       for(j=0;j<n-i-1;j++){
           if(strcmp(s[j],s[j+1])>0){
              t=s[j];
              s[j]=s[j+1];
              s[j+1]=t;
          }
       }
    }
}