关于C语言中字符指针的困惑 [英] Confusion about Character Pointers in C

问题描述

C语言中的字符指针使我感到困惑.

The character pointers in C is confusing me.

假设我们有一个char指针，它指向字符串 constant 的第一个字符.

Suppose we have a char pointer, which points to the first character of string constant.

char *a="ABCD";

然后我们不能使用指针a更改该字符的值，因为以下语句会导致分段错误.

Then we cannot change the value of that character using the pointer a, as following statement results in a segmentation fault.

*a='X';

现在假设我们有一个char指针，它指向一个字符 constant .

Now suppose we have a char pointer, which points to a character constant.

const char B='X';    
char *ptr=&B;

然后我们可以使用语句更改该字符的值

Then we are allowed to change the value of that character using the statement

*ptr='Z';

我的问题是，这是未定义行为的情况，证明C不够健壮吗?还是涉及更深层次的逻辑?

My question is that is this a case of undefined behaviour proving C is not robust? Or is there some deeper logic involved?

推荐答案

指针的行为有所不同的原因是C程序具有多个内存段，其中一些是受保护的.

The reason the pointer behaves differently is that C program has several memory segments, some of which are protected.

然后我们不能使用指针a更改该字符的值，因为以下语句会导致分段错误.

Then we cannot change the value of that character using the pointer a, as following statement results in a segmentation fault.

与其他常量不同，字符串常量被放置在受保护的段中.尝试修改此段会导致未定义的行为(即您的程序可能会出现段错误).

Unlike other constants, string constants are placed into a protected segment. Any attempt to modify this segment results in undefined behavior (i.e. your program could segfault).

由于指针指向字符串常量，因此不能修改其指向的值.如果使用数组将常量的副本强制放入可修改的内存中，则将允许相同的修改:

Since your pointer points into a string constant, the value to which it points cannot be modified. If you force a copy of the constant into modifiable memory by using an array, the same modification would be allowed:

char a[]="ABCD";
*a='X';

然后我们就可以使用以下语句更改该[单个]字符的值

Then we are allowed to change the value of that [single] character using the statement

这是因为字符常量总是复制到可修改的内存中.

This is because character constant is always copied into modifiable memory.

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