简单 C 代码中的分段错误(核心转储) [英] Segmentation fault (core dumped) in a simple C code

问题描述

我是 C 的新手.我指的是 Brian W Kernighian 和 Dennis Ritchie 所著的《The C Programming Language》一书.书中给出了指针递增和赋值的代码如下.

I am new in C. I am referring to the book "The C Programming Language" by Brian W Kernighian and Dennis Ritchie. There is a code for pointer increment and assignment given in the book as follows.

#include<stdio.h>

int main()
    {
        char *s = "Goal";
        char *t = "Home";
        while(*s++ = *t++) printf(*s);
        return 0;
    }

使用命令保存和编译代码

The code is saved and compiled using the command

gcc ptr.c -o ptr -std=c99

现在通过运行命令运行代码

Now on running the code by running command

./ptr

我收到以下错误

分段错误(核心转储)

错误似乎在 while 循环条件内.但代码与书中给出的完全相同.我错过了什么?

The error seems to be inside the while loop condition. But the code is exactly as given in the book. What am I missing?

推荐答案

s 和 t 都是字符串文字，不能修改字符串文字.但是这段代码

s and t are both string literals, and you can't modify a string literal. But this piece of code

*s++ = *t++

会修改s，导致分段错误.

will modify s, which causes segmentation fault.

要修复它，请使用 char 数组.我还修改了 printf 部分以使其合法.

To fix it, use a char array. I also modified the printf part to make it legal.

#include<stdio.h>

int main()
{
    char arr[] = "Goal";
    char *s = arr;
    char *t = "Home";
    while(*s++ = *t++) 
        ;
    printf("%s
", arr);
    return 0;
}

不过，我认为这个程序最好使用单独的函数来复制字符串，程序看起来会更清晰.

However, I think this program is better done using an individual function to copy the string, the program will look clearer.

#include<stdio.h>
void my_strcpy(char *s, char *t);

int main()
{
    char s[] = "Goal";
    char *t = "Home";
    my_strcpy(s, t);
    printf("%s
", s);
    return 0;
}

void my_strcpy(char *s, char *t)
{
    while(*s++ = *t++) 
        ;
}

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