分段错误扭转一个字符串字面量 [英] Segmentation fault reversing a string literal

问题描述

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;INT主要（无效）
{
    //个char [6] = {'H'，'E'，'L'，'L'，'O'，'\\ 0'};
    的char * s =你好;
    INT I = 0，M;
    焦温度;    INT N = strlen的（S）;
    // S [N] ='\\ 0';
    而（ⅰ≤（N / 2））
    {
         TEMP = *（S + I）; //使用空字符作为临时存储。
         *（S + I）= *（S + N-I-1）;
         *（S + N-I-1）=温度;
         我++;
    }
    的printf（转字符串=％s的\\ n，S）;
    系统（暂停）;
    返回0;
}
 

在编译错误的是段错误（访问冲突）。请告诉正是这两个定义之间的区别：

 个char [6] = {'H'，'E'，'L'，'L'，'O'，'\\ 0'};
的char * s =你好;
 

解决方案

您code尝试修改字符串文字是不是在C允许或C ++如果你改变：

 的char * s =你好;
 

到

 个char [] =你好;
 

然后要修改的阵列，成文字已被复制的内容（相当于初始化与单个字符数组），这是确定。

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    //char s[6] = {'h','e','l','l','o','\0'};
    char *s = "hello";       
    int i=0,m;
    char temp;

    int n = strlen(s);
    //s[n] = '\0';
    while (i<(n/2))
    {
         temp = *(s+i);       //uses the null character as the temporary storage.
         *(s+i) = *(s+n-i-1);
         *(s+n-i-1) = temp;
         i++;
    }
    printf("rev string = %s\n",s);
    system("PAUSE");
    return 0;
}

On the compilation the error is segmentation fault (access violation). Please tell what is the difference between the two definitions:

char s[6] = {'h','e','l','l','o','\0'};
char *s = "hello"; 

解决方案

Your code attempts to modify a string literal which is not allowed in C or C++ If you change:

char *s = "hello"; 

to:

char s[] = "hello"; 

then you are modifying the contents of the array, into which the literal has been copied (equivalent to initialising the array with individual characters), which is OK.

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