什么时候构造函数调用静态对象

Question

I understand that static variables are initialized at compile time, but what about a static object?

eg if I have the following code:

class A {
    A();
};

A::A(){
    std::cout << "Constructing A" << std::endl;
}

int main(){
    std::cout << "Hello World!" << std::endl;
    static A A_obj;
    std::cout << "Goodbye cruel world" << std::endl;
    return 0;
}

Should I expect the output to be:

Hello World!
Constructing A
Goodbye cruel world

or

Constructing A
Hello World!
Goodbye cruel world

Answer 1

"I understand that static variables are initialized at compile time"

Not true. static variables at function scope are strictly initialised at the point of first encounter. And will be destructed after the closing brace of main .

You'll get output in this order:

Hello World!
Constructing A
Goodbye cruel world

Answer 2

According to cppreference :

Variables declared at block scope with the specifier static have static storage duration but are initialized the first time control passes through their declaration (unless their initialization is zero- or constant-initialization, which can be performed before the block is first entered). On all further calls, the declaration is skipped.

Meaning the constructor is called as soon as the declaration is seen for the class. Furthermore:

The destructor for a block-scope static variable is called at program exit , but only if the initialization took place successfully.

Since the program exit is at the end of the function main() , once your program reaches the end of main() , the destructor of your class will be called.

Thus, you should expect the first output for the program.

Answer 3

I understand that static variables are initialized at compile time, but what about a static object?

That's an incorrect understanding.

Any variable, static or not, can be evaluated at compile time if that's possible. Some are required to be evaluated at compile time. Some cannot be evaluated at compile time, and hence must be initialized at run time. Once again, that does not depend on whether a variable is static or not.

Coming to your code, you should expect the first block of output.

The code to initialize A_obj is executed after the previous line has been executed.