在 C++ 中，用自身初始化全局变量是否具有未定义的行为？

Question

int i = i;

int main() { 
 int a = a;
 return 0;
} 

int a = a肯定有未定义的行为 (UB)，关于它的更多细节在 读取未初始化的值总是未定义的行为吗？或者有例外吗？.

但是呢int i = i？在 C++ 中，我们可以为全局变量分配非常量值。i在遇到声明之前声明并零初始化（因为它具有文件范围）。在这种情况下，我们将0在定义的后面分配给它。可以说这没有 UB 吗？

Answer 1

令人惊讶的是，这并不是未定义的行为。

静态初始化 [basic.start.static]

Constant initialization is performed if a variable or temporary object with static or thread storage duration is constant-initialized. If constant initialization is not performed, a variable with static storage duration or thread storage duration is zero-initialized. Together, zero-initialization and constant initialization are called static initialization; all other initialization is dynamic initialization. All static initialization strongly happens before any dynamic initialization.

Important parts bold-faced. "Static initialization" includes global variable initialization, "static storage duration" includes global variables, and the above clause is applicable here:

int i = i;

This is not constant-initialization. Therefore, zero-initialization is done according to the above clause (for basic integer types zero-initialization means, unsurprising, that it's set to 0). The above clause also specifies that zero initialization must take place before dynamic initialization.

So, what happens here:

1. i is initialized to 0.
2. i is then dynamically initialized, from itself, so it still remains 0.

Answer 2

该行为可能未定义i，因为根据您阅读标准的方式，您可能会i在其生命周期开始之前阅读。

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[basic.life]/1.2

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...类型 T 的对象的生命周期从以下时间开始：

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\xe2\x80\x94 其初始化（如果有）已完成...

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正如另一个答案中提到的，i初始化两次：首先静态零初始化，然后i动态初始化。

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哪个初始化开始生命周期？第一个还是最后一个？

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该标准含糊不清，并且其中存在相互矛盾的注释（尽管它们都是非规范性的）。[basic.life]/6首先， （感谢@eerorika ）中有一个脚注明确指出动态初始化启动生命周期：

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[basic.life]/6

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在对象的生命周期开始之前但在对象将占用的存储空间已分配之后²⁶

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...

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26) 例如，在动态初始化具有静态存储持续时间的对象之前...

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这种解释对我来说最有意义，因为否则在类实例进行动态初始化、建立其不变量（包括标准定义的标准库类）之前访问类实例是合法的。

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中还有一个相互矛盾的注释[basic.start.static]/3，但该注释比我上面提到的更早。

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