实验五
任务一
1.1
#include <stdio.h> #define N 4 int main() { int x[N] = { 1,9,8,4 }; int i; int* p; // 方式1：通过数组名和下标遍历输出数组元素 for (i = 0; i < N; ++i) printf("%d", x[i]); printf("\n"); // 方式2：通过指针变量遍历输出数组元素 （写法1） for (p = x; p < x + N; ++p) printf("%d", *p); printf("\n"); // 方式2：通过指针变量遍历输出数组元素（写法2） p = x; for (i = 0; i < N; ++i) printf("%d", *(p + i)); printf("\n"); // 方式2：通过指针变量遍历输出数组元素（写法3） p = x; for (i = 0; i < N; ++i) printf("%d", p[i]); printf("\n"); return 0; }
1.2
#include<stdio.h> int main() { int x[2][4] = {{1,9,8,4},{2,0,4,9}}; int i,j; int *p; int(*q)[4]; for (i = 0;i<N;++i) { for (j = 0;j<4;++i) printf("%d",x[i][j]); printf("\n"); } for (p = &x[0][0],i = 0;p < &x[0][0] + 8;++p,++i) { printf("%d", *p); if ((i + 1) % 4 == 0) printf("\n"); } for (q = x; q < x + 2; ++q) { for (j = 0; j < 4; ++j) printf("%d", *(*q + j)); printf("\n"); } return 0; }
任务二
2.1
#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大小是24，sizeof统计的是占用内存大小，包含休止符，strlen统计的是字符串长度，不包含休止符。
不能，等号两边类型不同
对
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; 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; }
sizeof计算的是内存大小，strlen计算的是字符串长度。
可以，1是地址常量，2是指针变量。
交换s1和s2的指向关系，储存内容不变。
任务三
#include <stdio.h> void str_cpy(char *target, const char *source); void str_cat(char *str1, char *str2); int main() { char s1[80], s2[20] = "1984"; str_cpy(s1, s2); puts(s1); str_cat(s1, " Animal Farm"); puts(s1); return 0; } void str_cpy(char *target, const char *source) { while (*target++ = *source++) ; } void str_cat(char *str1, char *str2) { while (*str1) str1++; while (*str1++ = *str2++) ; }
任务四
#include <stdio.h> #define N 80 int func(char *); int main() { char str[80]; while (gets(str) != NULL) { if (func(str)) printf("yes\n"); else printf("no\n"); } return 0; } int func(char *str) { char *begin, *end; begin = end = str; while (*end) end++; end--; while (begin < end) { if (*begin != *end) return 0; else { begin++; end--; } } return 1; }
任务五
#include <stdio.h> #define N 80 void func(char *); int main() { char s[N]; while (scanf("%s", s) != EOF) { func(s); puts(s); } return 0; } void func(char *str) { int i; char *p1, *p2, *p; p1 = str; while (*p1 == '*') p1++; p2 = str; while (*p2) p2++; p2--; while (*p2 == '*') p2--; p = str; i = 0; while (p < p1) { str[i] = *p; p++; i++; } while (p <= p2) { if (*p != '*') { str[i] = *p; i++; } p++; } while (*p != '\0') { str[i] = *p; p++; i++; } str[i] = '\0'; }
任务六
6.1
#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; } }
6.2
#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; char *tmp; for (i = 0; i < n - 1; ++i) { k=i; for (j = i +1; j < n;j++) if (strcmp(name[j], name[j + 1]) > 0) k=j; if (k!=i) { tmp = name[j]; name[j] = name[j + 1]; name[j + 1] = tmp; } } }
变换的是指针变量，内存不变
任务七
#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, j, k; char* tmp; tmp=str; while ((*tmp <='9' && *tmp >= '0') || *tmp == 'X') tmp++; if (*tmp == '\0' && strlen(str) == 18) return 1; else return 0; }
任务八
#include <stdio.h> #define N 80 void encoder(char* s); // 函数声明 void decoder(char* s); // 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* s) { char* tmp; tmp = s; int i = 0; while (*tmp) { if (*tmp == 'z') s[i++] = 'a', tmp++; else if (*tmp == 'Z') s[i++] = 'A', tmp++; else if (*tmp < 'A' || *tmp>'z') s[i] = s[i++], tmp++; else s[i++] = *tmp + 1,tmp++; } } void decoder(char* s) { char* tmp; tmp = s; int i = 0; while (*tmp) { if (*tmp == 'a') s[i++] = 'z', tmp++; else if (*tmp == 'A') s[i++] = 'Z', tmp++; else if (*tmp < 'A' || *tmp>'z') s[i] = s[i++], tmp++; else s[i++] = *tmp - 1, tmp++; } }
