实验5
#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 = 1; i < n; ++i)
        if(x[i] < *pmin)
            *pmin = x[i];
        else if(x[i] > *pmax)
            *pmax = x[i];
}#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 = 1; i < n; ++i)
        if(x[i] < *pmin)
            *pmin = x[i];
        else if(x[i] > *pmax)
            *pmax = x[i];
}

找出数组中的最大值和最小值
#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 = 1; i < n; ++i)
        if(x[i] > x[max_index])
            max_index = i;
    
    return &x[max_index];
}

找出数组中的最大值,并返回最大值的地址
#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[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;
}#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[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;

1,数组s1的大小是4, sizeof(s1)计算的是字符串所占用的字节数,strlen(s1) 统计的是有效字符数,不包括'\0'
2,不能替换成以下写法,s1是字符串数组的起始地址常量,不能表示一个数组。
#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;
}

1,指针变量s1中存放的是字符串的初始地址 , sizeof(s1) 计算的是指针变量s1所占用的字节数, strlen(s1) 统计的是s1所指向的字符串所占用的有效字节数
2,line6代码能替换成下面的写法,对比task2_1.c中的line7, task2_1中s1是地址常量,不能复制,task2_2中s1是指针变量,该语义是把字符串的初始地址赋值给
3,line19-line21,交换的是地址,s1和s2所指向的字符串改变,字符串常量在内存中的存储单元没有交换。
#include <stdio.h>
#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: 使用指向元素的指针变量ptr1间接访问二维数组元素\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: 使用指向一维数组的指针变量ptr2间接访问二维数组元素\n");
    for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
        for (j = 0; j < 4; ++j)
            printf("%d ", *(*ptr2 + j));
        printf("\n");
    }
    return 0;
}#include <stdio.h>
#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: 使用指向元素的指针变量ptr1间接访问二维数组元素\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: 使用指向一维数组的指针变量ptr2间接访问二维数组元素\n");
    for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
        for (j = 0; j < 4; ++j)
            printf("%d ", *(*ptr2 + j));
        printf("\n");
    }
    return 0;
}

#include<stdio.h>
#define N 80
void replace(char *str,char old_char,char new_char);
int main()
{
    char text[N]="c programming is difficult or not,it is a question.";
    printf("原始文本:\n");
    printf("%s",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;
}
void replace(char *str,char old_char,char new_char)
{
    while(*str)
    {
        if(*str==old_char)
            *str=new_char;
        *str++;
    }
}

把old_char替换成new_char
可以
#include <stdio.h>
#define N 80
void str_trunc(char *str, char x);
int main() {
    char str[N];
    char ch;
    printf("输入字符串: ");
    gets(str);
    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", str);
}
void str_trunc(char *str, char x) {
    while(*str) {
        if(*str == x)
          *str='\0';     // blank1
 			str++;  // blank2
    }
      str--;    // blank3

#include<stdio.h>
#include<string.h>
void sort(char *name[],int n);
int main()
{
    char *course[4]={"C Program",
                    "C++ Object Oriented Program",
                    "Operating System",
                    "Data Structure and Algorithms"};
    int i;
    sort(course,4);
    for(i=0;i<4;++i)
        printf("%s\n",course[i]);
    return 0;
}
void sort(char *name[],int n)
{
    int i,j;
    char *tmp;
    for(i=0;i<n-1;++i)
        for(j=0;j<n-1-i;++j)
            if(strcmp(name[j],name[j+1])>0)
            {
                tmp=name[j];
                name[j]=name[j+1];
                name[j+1]=tmp;
            }
}

#include <stdio.h>
#include <string.h>
void sort(char *name[], int n);
int main() {
    char *course[4] = {"C Program",
                       "C++ Object Oriented Program",
                       "Operating System",
                       "Data Structure and Algorithms"};
    int i;
    sort(course, 4);
    for (i = 0; i < 4; i++)
        printf("%s\n", course[i]);
    return 0;
}
void sort(char *name[], int n) {
    int i, j, k;
    char *tmp;
    for (i = 0; i < n - 1; i++) {
        k = i;
        for (j = i + 1; j < n; j++)
            if (strcmp(name[j], name[k]) < 0)
                k = j;
        if (k != i) {
            tmp = name[i];
            name[i] = name[k];
            name[k] = tmp;
        }
    }
}

task6
#include <stdio.h>
#include <string.h>
#define N 5
int check_id(char *str); // 函数声明
int main() {
    char *pid[N] = {"31010120000721656X",
                    "330106199609203301",
                    "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;
}
// 函数定义
// 功能: 检查指针str指向的身份证号码串形式上是否合法。
// 形式合法,返回1,否则,返回0
int check_id(char *str) {
		int k;
	    k=strlen(str);
		if(k!=18)
		return 0;
		
		for(k=0;k<18;k++){
			if(k==17){
				if(*(str+k)>'9'||*(str+k)<'0'&&*(str+k)!='X')
				return 0;
			}
			else if(*(str+k)>'9'||*(str+k)<'0')
			return 0;
		}
			
}

task7
#include <stdio.h>
#define N 80
void encoder(char *str); 
void decoder(char *str); 
int main() {
    char words[N];
    printf("输入英文文本: ");
    gets(words);
    printf("编码后的英文文本: ");
    encoder(words); 
    printf("%s\n", words);
    printf("对编码后的英文文本解码: ");
    decoder(words); 
    printf("%s\n", words);
    return 0;
}
void encoder(char *str) {
    while(*str) {
        if((*str>='a'&&*str<'z')||(*str>='A'&&*str<'Z'))
    (*str)++;
    else if(*str=='z')
    *str='a';
    else if(*str=='Z')
    *str='A';
    str++;
}
    
    }
  
void decoder(char *str) {
    while(*str){
    if((*str>'a'&&*str<'z')||(*str>='A'&&*str<'Z'))
    (*str)--;
    else if(*str=='a')
    *str='z';
    else if(*str=='A')
    *str='Z';
    str++;
}
 
}



                    
                
                
            
        
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