为什么 char* 会导致未定义的行为，而 char[] 不会? [英] Why does char* cause undefined behaviour while char[] doesn't?

问题描述

尝试修改字符串文字会导致未定义的行为:

Attempting to modify a string literal causes undefined behavior:

char * p = "wikipedia"; 
p[0] = 'W'; // undefined behaviour

防止这种情况的一种方法是将其定义为数组而不是指针:

One way to prevent this is defining it as an array instead of a pointer:

char p[] = "wikipedia"; 
p[0] = 'W'; // ok

为什么 char* 会导致未定义的行为，而 char[] 不会?

Why does char* cause undefined behaviour, while char[] doesn't?

推荐答案

任何修改 C 字符串文字的尝试都有未定义的行为.编译器可能会安排将字符串文字存储在只读内存中(受操作系统保护，而不是字面意义上的 ROM，除非您在嵌入式系统上).但是语言不需要这个；作为一名程序员，这取决于您是否正确.

Any attempt to modify a C string literal has undefined behaviour. A compiler may arrange for string literals to be stored in read-only memory (protected by the OS, not literally ROM unless you're on an embedded system). But the language doesn't require this; it's up to you as a programmer to get it right.

一个足够聪明的编译器可能会警告您，您应该将指针声明为:

A sufficiently clever compiler could have warned you that you should have declared the pointer as:

const char * p = "wikimedia";

虽然没有 const 的声明在 C 中是合法的(为了不破坏旧代码).但是不管有没有编译器警告，const 都是一个非常好的主意.

though the declaration without the const is legal in C (for the sake of not breaking old code). But with or without a compiler warning, the const is a very good idea.

(在 C++ 中，规则不同；C++ 字符串文字，与 C 字符串文字不同，实际上是 const.)

(In C++, the rules are different; C++ string literals, unlike C string literals, really are const.)

当您使用文字初始化数组时，文字本身仍然存在于程序映像的潜在只读区域中，但它被复制到本地数组中:

When you initialize an array with a literal, the literal itself still exists in a potentially read-only region of your program image, but it is copied into the local array:

char s[] = "wikimedia"; /* initializes the array with the bytes from the string */
char t[] = { 'w', 'i', ... 'a', 0 };  /* same thing */

_{请注意，char u[] = *p 不 起作用——数组只能从大括号初始化器中初始化, 和另外来自字符串文字的 char 数组.}

_{Note that char u[] = *p does not work -- arrays can only be initialized from a brace initializer, and char arrays additionally from a string literal.}

这篇关于为什么 char* 会导致未定义的行为，而 char[] 不会?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！