在标头中声明全局变量“extern const int"，但在源文件中仅声明“int" [英] Declaring a global variable `extern const int` in header but only `int` in source file

问题描述

我在用 GCC 做实验，发现你可以在头文件中声明外部变量 const 但在实现文件中保持可变.

I was experimenting with GCC and found out that you can declare external variables const in header files but keep them mutable in implementation files.

编辑:这实际上不起作用.我编译测试代码的唯一原因是我没有在header.c"中包含header.h".

EDIT: This does actually not work. The only reason I got my test code to compile was because I did not include "header.h" in "header.c".

header.h:

#ifndef HEADER_H_
#define HEADER_H_

extern const int global_variable;

#endif

header.c:

int global_variable = 17;

这似乎是一个非常好的功能，用于保持 global_variable 对 header.h 的用户只读，但保持它们可通过实现(header.h)修改.c).

This seems like a very good feature to use for keeping global_variable readonly to the users of header.h but keeping them modifable by the implementation (header.c).

注意:以下代码只是说明这种声明方式如何防止分配给 global_variable 的示例.

NOTE: The following code is just an example on how this way of declaring will prevent assignment to global_variable.

#include "header.h"

int main(void)
{
    global_variable = 34; /* This is an error as `global_variable` is declared const */
    return 0;
}

因为我以前从未在实践中看到过技术.我开始怀疑它是否有效.

Because I have never seen technique in practise before. I start to wonder if it is valid.

这是明确定义的行为还是 GCC 未能警告我的错误?

Is this well defined behaivor or is this an error that GCC fails to warn me about?

推荐答案

const int 和 int 不是兼容的类型.

const int and int are not compatible types.

例如:

extern const int a;

int a = 2;

在 C 中无效，正如 C 所说:

is not valid in C as C says that:

(C11, 6.7p4) "同一范围内引用同一对象或函数的所有声明应指定兼容类型"

(C11, 6.7p4) "All declarations in the same scope that refer to the same object or function shall specify compatible types"

在您的情况下，它们不在同一范围内(不同的翻译单元)，但 C 也表示:

In your case they are not in the same scope (different translation units) but C also says that:

(C11, 6.2.7p2) "所有引用同一对象或函数的声明都应具有兼容的类型；否则，行为未定义."

(C11, 6.2.7p2) "All declarations that refer to the same object or function shall have compatible type; otherwise, the behavior is undefined."

由于您违反了上述规则，因此您正在调用未定义的行为.

As you are violating the rule above, you are invoking undefined behavior.

请注意，C90 有相同的段落.

Note that C90 has the same paragraphs.

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