静态变量与自身的初始化

问题描述

请考虑以下代码示例：

#include <iostream>

static int bar = bar;

int main()
{
    int foo = foo;
    std::cout << "foo = " << foo << std::endl;
    std::cout << "bar = " << bar << std::endl;
}

我得到以下编译器警告：

I get following compiler warning:

main.cpp: In function 'int main()':
main.cpp:7:15: warning: 'foo' is used uninitialized in this function [-Wuninitialized]
     int foo = foo;
               ^

输出：

foo = 0
bar = 0

期望这个警告作为foo使用单元化。而不是0，'foo'可以是任何东西。
自定义未定义。

I expected this warning as foo is used unitialised. Instead of 0, 'foo' can be anything.Self-assignment is undefined.

但是为什么'bar'的自我赋值不会被警告呢？这是'bar'定义或未定义的行为的分配和为什么？

But why is the the self-assignment of 'bar' not warned? Is this assignment of 'bar' defined or undefined behaviour and why?

我知道，元素数据类型的静态变量初始化为'0'，但在这种情况下，在其初始化期间使用变量bar。我想知道，如果这是定义的行为和'0'是预期的输出。 （这将解释，没有编译器警告发生）。

I know, static variables of elementar data types are initialised with '0' but in this case, the variable 'bar' is used during its initialisation. I'm wondering, if this is defined behaviour and the '0' is the expected output. (Which would explain, that no compiler warning occurs).

推荐答案

我相信标准的一部分与您的问题相关（§3.6.2/ 2）：

I believe that part of the standard is relevant to your question (§3.6.2/2):

所以在这种情况下，甚至在编译之前看看你的定义 bar ，它已经零初始化了它。

So in this case, even before the compiler looks at your definition of bar, it has already zero-initialized it.

在标准中，静态变量初始化应该有两个阶段（强调我的）。

As it is specified a bit further in the standard, there shall be two phases for static variables initialization (emphasis mine).

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