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_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];
}

ans1:找到最大值和最小值。
ans2：x【0】所在的地址。
taks1.2

#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];

ans1:找到最大值。
ans2: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;
}

ans1:80;数组在内存中的占用的字节总数；统计字符串中字符的个数。
ans2:不可以，s1是数组名，是一个常量指针，不能用字符串赋值。
ans3：会交换。

#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;
}

ans1：字符串常量的地址；指针本身在内存中的占用情况；指针指向的字符串的长度。
ans2：可以，2.1中的是数组，2.2中是将字符串的地址赋值给s1。
ans3：交换的是s1与s2中的指针常量，两个字符串常量的内存位置并未交换。
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;
}

ans1：int (*ptr)[4];是个数组指针，指针指向的是数组。
ans2：int *ptr[4];是个指针数组，数组中存放的是指针。

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++;
    }
}

ans1：将一种字符替换为另一种。
ans2：不可以。
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;
}
char* str_trunc(char* str, char x) {
    for (int i = 0; i < N; i++) {
        if (*str == x) {
            *str = '\0';
            break;
        }

        str++;
    }


}

ans：解决循环输入中回车的影响。
task6

#include <stdio.h>
#include <string.h>
#define N 5
#include<ctype.h>
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) {
    if (strlen(str) != 18) {
        return 0;
    }
    for (int i = 0; i < 17; i++) {
        if(!isdigit((unsigned char)str[i]))
        return 0;
    }
    char last = str[17];
    if (!isdigit((unsigned char)last) && last != 'X')
    {
        return 0;
    }
    return 1;

}

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_s("%d", &n);
	printf("编码后的英文文本: ");
	encoder(words, n);
	printf("%s\n", words);
	// 函数调用
	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] - 'a' + n + 26) % 26;
		}
		else if (str[i] >= 'A' && str[i] <= 'Z') {
			str[i] = 'A' + (str[i] - 'A' + n + 26) % 26;
		}
		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] - 'a' - n + 26) % 26;
		}
		else if (str[i] >= 'A' && str[i] <= 'Z') {
			str[i] = 'A' + (str[i] - 'A' - n + 26) % 26;
		}
		i++;
	}
}

task8

#include <stdio.h>
int main(int argc, char *argv[]) {
int i;
for(i = 1; i < argc; ++i)
printf("hello, %s\n", argv[i]);
return 0;
}

#include <stdio.h>
#include <string.h>
int main(int argc, char* argv[]) {
	int i;
	for (int j=1; j < argc-1 ;j++) {
		for (int m = 1; m < argc-j; m++) {
			if (strcmp(argv[m], argv[m + 1]) > 0) {
				char *temp = argv[m];
				argv[m] = argv[m + 1];
				argv[m + 1] = temp;
			
			}

		}
	}
	for (i = 1; i < argc; ++i)
		printf("hello, %s\n", argv[i]);
	return 0;
}