通过替换字符串的字母在C中进行分段错误 [英] segmentation fault in C by replacing a string's letter

问题描述

我想从字符串中替换字符（例如第二个字符）。
我的代码有什么问题？它可以编译，但是没有执行我需要的操作，而是给我一个Segmentation Fault。谢谢！

I want to replace a character (for example the second one) from a string. What's wrong with my code? It can compile, but instead of doing what I need it give me a Segmentation fault. Thanks!

#define _XOPEN_SOURCE
#include <unistd.h>
#include <stdio.h>
#include <cs50.h>
#include <string.h>

int main (int argc, string argv[])
{
    string key="abcd";
    key[1]='f';
}

在编译我的代码后

~/workspace/pset2/crack/ $ clang -ggdb3 -O0 -std=c11 -Wall -Werror -Wshadow bug.c -lcrypt -lcs50 -lm -o bug
~/workspace/pset2/crack/ $ ./bug
Segmentation fault

推荐答案

您只能修改堆或堆栈中的内存，您的 abcd 称为字符串文字，属于程序的只读内存空间，因此不能在其中进行修改（无论如何，它都不能在任何合适的操作系统下运行）。

You can only modify memory that is in the heap or stack, your "abcd" there is what is known as a String Literal and belongs in the program's read-only memory space, it cannot be modified where it is (not if it's running under any decent operational system anyway).

放置在程序中的原因首先是因为 string 基本上只是一个 char * 指针，它会指向任何内存空间，并且不会不在乎它是否位于只读空间中。 （不要与C ++ std :: string 混淆）。

The reason it's placed there to begin with is because string is basically just a char * pointer, and it will point to any memory space and doesn't care if it's in read-only space or not. (not to be confused with C++ std::string).

要使其可修改，您必须告诉C

To make it modifiable you have to tell C it's not a pointer, but an array.

char key[] = "abcd";

现在C将把此字符串放在堆栈中，您可以随意对其进行修改。

Now C will place this string in the stack where you can modify it at will.

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