更改一个c字符串中的一个字符

问题描述

我试图了解以下代码为何非法:

I am trying to understand why the following code is illegal:

int main ()
{
    char *c = "hello";
    c[3] = 'g'; // segmentation fault here
    return 0;
}

编译器遇到char *c = "hello";时会做什么?

What is the compiler doing when it encounters char *c = "hello";?

按照我的理解，它是char的自动数组，而c是第一个char的指针.如果是这样，c[3]就像*(c + 3)一样，我应该可以进行分配.

The way I understand it, its an automatic array of char, and c is a pointer to the first char. If so, c[3] is like *(c + 3) and I should be able to make the assignment.

只是试图了解编译器的工作方式.

Just trying to understand the way the compiler works.

推荐答案

字符串常量是不可变的.即使将它们分配给char *，也无法更改它们(因此，将它们分配给const char *，这样就不会忘记).

String constants are immutable. You cannot change them, even if you assign them to a char * (so assign them to a const char * so you don't forget).

要更详细些，您的代码大致等同于:

To go into some more detail, your code is roughly equivalent to:

int main() {
  static const char ___internal_string[] = "hello";
  char *c = (char *)___internal_string;
  c[3] = 'g';
  return 0;
}

此___internal_string通常分配给只读数据段-任何在此更改数据的尝试都将导致崩溃(严格来说，也可能发生其他结果-这是未定义行为"的一个示例) .但是，由于历史原因，编译器允许您分配给char *，给您一种错误的印象，您可以对其进行修改.

This ___internal_string is often allocated to a read-only data segment - any attempt to change the data there results in a crash (strictly speaking, other results can happen as well - this is an example of 'undefined behavior'). Due to historical reasons, however, the compiler lets you assign to a char *, giving you the false impression that you can modify it.

请注意，如果执行此操作，它将起作用:

Note that if you did this, it would work:

char c[] = "hello";
c[3] = 'g'; // ok

这是因为我们正在初始化一个非常量字符数组.尽管语法看起来很相似，但是编译器对它的处理方式有所不同.

This is because we're initializing a non-const character array. Although the syntax looks similar, it is treated differently by the compiler.

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