实验五
task1_1.c
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#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];
}
问题1：在数组x中找出最小值和最大值
问题2：指向min和max变量的地址
task1_2.c
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#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_s("%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];
}
问题1：在数组x中找出最大值并返回最大值所在元素的地址
问题2：可以
task2_1.c
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#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;
}
task2_2.c
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#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;
}
问题1：字符串常量“Learning makes me happy”在内存中的首地址；指针变量 s1 本身所占的内存字节数；“Learning makes me happy”中字符的数量。
问题2：可以
问题3：指针变量 s1 和 s2 的值；否
task3.c
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#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;
}
task4.c
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#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++;
}
}
问题1：将字符串 str中与 old_char 相等的字符，替换为 new_char
问题2：可以
task5.c
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#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* p = str;
while (*p != '\0') {
if (*p == x) {
*p = '\0';
break;
}
p++;
}
return str;
}
问题：读取不到完整的字符串；清楚输入缓冲区中的换行符
task6.c
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#include <stdio.h>
#include <string.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 len = strlen(str);
if (len != 18) {
return 0;
}
int i;
for (i = 0; i < len - 1; i++) {
if (str[i] < '0' || str[i] > '9') {
return 0;
}
}
if (str[len - 1] != 'X' && (str[len - 1] < '0' || str[len - 1] > '9')) {
return 0;
}
return 1;
}
task7.c
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#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_s("%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 + 'a';
}
else if ((*str >= 'A' && *str <= 'Z')) {
*str = (*str - 'A' + n) % 26 + 'A';
}
str++;
}
}
void decoder(char* str, int n) {
while (*str) {
if ((*str >= 'a' && *str <= 'z')) {
*str = (*str - 'a' - n + 26) % 26 + 'a';
}
else if ((*str >= 'A' && *str <= 'Z')) {
*str = (*str - 'A' - n + 26) % 26 + 'A';
}
str++;
}
}
task8.c
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