C++ 函数中静态变量的生命周期是多长？

如果一个变量被声明为static在函数的作用域中，它仅初始化一次并在函数调用之间保留其值。它的寿命究竟是多少？它的构造函数和析构函数什么时候被调用？

void foo() 
{ 
    static string plonk = "When will I die?";
}

The lifetime of function static variables begins the first time^[0] the program flow encounters the declaration and it ends at program termination. This means that the run-time must perform some book keeping in order to destruct it only if it was actually constructed.

Additionally, since the standard says that the destructors of static objects must run in the reverse order of the completion of their construction^[1], and the order of construction may depend on the specific program run, the order of construction must be taken into account.

Example

struct emitter {
    string str;
    emitter(const string& s) : str(s) { cout << "Created " << str << endl; }
    ~emitter() { cout << "Destroyed " << str << endl; }
};

void foo(bool skip_first) 
{
    if (!skip_first)
        static emitter a("in if");
    static emitter b("in foo");
}

int main(int argc, char*[])
{
    foo(argc != 2);
    if (argc == 3)
        foo(false);
}

Output:

C:>样本.exe
创建于 foo
在 foo 中被破坏

C:>样本.exe 1
创建于 if
创建于 foo
在 foo 中被破坏
毁于如果

C:>样本.exe 1 2
创建于 foo
创建于 if
毁于如果
在 foo 中被破坏

[0] Since C++98^[2] has no reference to multiple threads how this will be behave in a multi-threaded environment is unspecified, and can be problematic as Roddy mentions.

[1] C++98部分3.6.3.1 [基本.开始.术语]

[2]在 C++11 中，静态变量以线程安全的方式初始化，这也称为魔法静力学.