The C Programming Language 练习题2-5
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题目
编写函数 any(s1, s2),将字符串s2中的任一字符在字符串s1中第一次出现的位置作为结果返回。如果s1中不包含s2中的字符,则返回-1。(标准库函数strpbrk具有同样的功能,但它返回的是指向该位置的指针。)
题目分析
不太理解任一字符如何操作,就把每个字符在字符串1中的位置都做标注。实现方法跟2-4类似。
编程实现
#include <stdio.h>#define MAXLINE 1000int positionc(char s1[], char s2[], int p[]);int main(){ int i, positions[MAXLINE]; char c, sfirst[MAXLINE], ssecond[MAXLINE]; i = 0; printf("Please input string1:"); while ((c = getchar()) != '\n') sfirst[i++] = c; sfirst[i] = '\0'; i = 0; printf("Please input string2:"); while ((c = getchar()) != '\n') ssecond[i++] = c; ssecond[i] = '\0'; positionc(sfirst, ssecond, positions); i = 0; while (ssecond[i] != '\0') { printf("%c\t%d\n",ssecond[i], positions[i]); i++; }}int positionc(char s1[], char s2[], int p[]){ int m, n, l, r; n = 0; while (s2[n] != '\0') { p[n] = 0; m = l = 0; while (s1[m] != '\0') { if (s1[m] != s2[n] && s1[m+1] != '\0') l++; else if ( s1[m] == s2[n]) { p[n] = l; break; } else if (s1[m] != s2[n] && s1[m + 1] == '\0') { p[n] = -1; break; } m++; } if (s1[m] == '\0' && l == 0) p[n] = -1; n++; }}
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