用C以外的主要变量的作用域 [英] Scope of a variable outside main in C

问题描述

考虑code：

#include <stdio.h>

int x;

int main (void) 
{ }

X 是 0 在主的值。但是，这是为什么？我还没有宣布它为静态。或者是它假定静态，因为它是一个功能之外？

The value of x is 0 inside main. But why is that? I have not declared it to be static. Or is it assumed static as it is outside a function?

如果上面是真实的，它是如何使它从的extern不同？

If the above is true, how does it make it different from an extern?

推荐答案

这是既不静态也不的extern 。这对编译单元它在一个变量可见的，并且还将会从申报 X 是一个的extern 变量。

It's neither static nor extern. It's a variable visible for the compilation unit it's in, and additionally will be visible from all compilation units that declare x to be an extern variable.

为什么我说这是既不静态也不的extern ？

Why am I saying it's neither static nor extern?

如果是的extern ，那么，就必须以不同的编译单元X 声明就可以了。显然，这是你唯一的编译单元。

If it was extern, then, there must be a different compilation unit with x declaration on it. Clearly this is your only compilation unit.

如果是静态那么，没有的extern 引用将被允许 X 在这个编译单元定义的变量。我们知道，我们可以很容易地声明的extern 变量此 X 宣布在这里。

If it was static then, no extern reference would be allowed to x variable defined in this compilation unit. We know that we could easily declare an extern variable to this x declared here.

为什么 0 分配到 X ？因为，在 C ，所有全局变量初始化为 0 。它说，所以在C99标准6.7.8（10）。

Why is 0 assigned to x? Because, in C, all global variables initialize to 0. It says so in 6.7.8 (10) of the C99 standard.

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