#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]; }
find_max_min 的功能是在输入的5个数里面找出最大与最小的数字。
指向x[0],即x[]的第一个元素
task 1_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* ptr = &x[0]; int i; for (i = 1; i < n; ++i) if (x[i] > *ptr) ptr = &x[i]; return ptr; }
功能:找出5个数据中最大的。
可以。
task 2—2
#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; }
s1的大小是24
sizeof(s1)计算的是算上字符串0/的总长度
strlen(s1)统计的是不算0/的长度
不能,因为形式错误
交换了。
task 3
#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; }
task 4_1
#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] = "c 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++; } }
replace功能:将字符串中算有的‘i’替换成‘*’
可以。
task4_2
#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'; ++str; } printf("%s",str); }
#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; } }
task 5_2
#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; } } }
task 6
#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 i; if(strlen(str)!=18){ return 0; } for(i=0;i<17;++i){ if(str[i]<'0'||str[i]>'9'){ return 0; } } if(str[17]<'0'||(str[17]>'9'&&str[17]<'X')||str[17]>'X'){ return 0; } }
task 7
#include <stdio.h> #define N 80 void encoder(char *str); // 函数声明 void decoder(char *str); // 函数声明 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); // 函数调用 printf("%s\n", words); printf("对编码后的英文文本解码: "); decoder(words); // 函数调用 printf("%s\n", words); return 0; } /*函数定义 功能:对s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,用其后的字符替换; 其中,z用a替换,Z用A替换 其它非字母字符,保持不变 */ void encoder(char *str) { int i=0; while(str[i]!='\0'){ if(str[i]>='a'&&str[i]<'z'||str[i]>='A'&&str[i]<'Z'){ str[i]++; } else if(str[i]=='z'||str[i]=='Z'){ str[i]-=25; } ++i; } } /*函数定义 功能:对s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,用其前面的字符替换; 其中,a用z替换,A用Z替换 其它非字母字符,保持不变 */ void decoder(char *str) { int i=0; while(str[i]!='\0'){ if(str[i]>'a'&&str[i]<='z'||str[i]>'A'&&str[i]<='Z'){ str[i]--; } else if(str[i]=='a'||str[i]=='z'){ str[i]+=25; } ++i; } }
task 8
#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; }
标签:int,s1,char,实验,str,printf,void From: https://www.cnblogs.com/p2kp2k/p/17860291.html