如果在全局对象的析构函数中调用 std::exit 会发生什么？

Question

考虑以下代码：

#include <cstdlib>
struct Foo {
    ~Foo() {
        std::exit(0);
    }
} foo;
int main() {
}

它在我的Linux（GCC，Clang）和Windows（Visual Studio）上成功编译并以零终止。然而，当用MSYS2的GCC在Windows(g++ (Rev2, Built by MSYS2 project) 10.3.0)上编译时，它进入无限递归并死于堆栈溢出。这可以通过在 std::exit() 之前添加一些调试输出来检查；我一开始没有加上，以免想到std::cout.

的破坏

是否有任何 C++ 标准对这种行为有任何规定？是 well-defined/implementation-defined/undefined/etc 吗？

例如，最近的一些草稿 [support.start.term]/9.1 对 std::exit 的行为进行了如下描述：

First, objects with thread storage duration and associated with the current thread are destroyed. Next, objects with static storage duration are destroyed and functions registered by calling atexit are called. See [basic.start.term] for the order of destructions and calls.

指的是[basic.start.term]/1，我猜：

Constructed objects ([dcl.init]) with static storage duration are destroyed and functions registered with std::atexit are called as part of a call to std::exit ([support.start.term]). The call to std::exit is sequenced before the destructions and the registered functions.

我没有看到在析构函数中调用 std::exit 有任何直接限制。

旁注：请不要评论“这段代码不好”、“thou shalt not use destructors in global objects”（您可能不应该这样做）以及在评论中探讨 XY 问题。考虑这是一个好奇的学生提出的学术问题，他知道对他们的原始问题有更好的解决方案，但在探索 C++ 的广阔草地时偶然发现了这个怪癖。

Answer 1

[basic.start.main]/4:

If std::exit is called to end a program during the destruction of an object with static or thread storage duration, the program has undefined behavior.

如果在全局对象的析构函数中调用 std::exit 会发生什么？

What should happen if one calls `std::exit` in a global object's destructor?

c++

language-lawyer

destructor

terminate