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所有这一切都试图做的是将最短字符串的地址放在数组的开头,将最长字符串的地址放在数组的末尾,但我看不出它有什么问题。当我在 linux 上运行它时,出现“分段错误”

#include <stdio.h>
#include <string.h>

void fx(char* t[], int n);

int main(void) {
    char*t[]= {"horse", "elephant", "cat", "rabbit"};
    int n, i;
    n = sizeof(t)/sizeof(t[0]);
    fx(t, n);
    printf("shortest is %s, longest is %s\n", t[0], t[n-1]);

}

void fx(char* t[], int n) {
    int i;
    char* temp, len0, len1, len;
    len0 = strlen(t[0]); /* lenght of first string*/
    len1 = strlen(t[n+1]); /*lenght of last string*/
    for(i=0; i<n; i++) {
        len = strlen(t[i]); /*temporary lenght if ith string*/
        if( len < len0 ) {
            temp = t[0];  /* if shorter, swap places with first*/
            t[0] = t[i];
            t[i] = temp;
        }
        else if(len > len1) {  /* if larger, swap places with last*/
            temp = t[n-1]; 
            t[n-1] = t[i];
            t[i] = temp;
        }
    }
}
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1 回答 1

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数组是 [0....n-1]

len1 = strlen(t[n+1]); /*lenght of last string*/

应该

len1 = strlen(t[n-1]); /*lenght of last string*/

len0、len1 和 len 必须是 int,而不是 char*,因为它们包含字符串的 len

于 2013-03-21T19:20:06.760 回答