C++ destruction of temporary object in an expressi

问题:

Given the following code:

#include <iostream>

struct implicit_t
{
    implicit_t(int x) :
        x_m(x)
    {
        std::cout << "ctor" << std::endl;
    }

    ~implicit_t()
    {
        std::cout << "dtor" << std::endl;
    }

    int x_m;
};

std::ostream& operator<<(std::ostream& s, const implicit_t& x)
{
    return s << x.x_m;
}

const implicit_t& f(const implicit_t& x)
{
    return x;
}

int main()
{
    std::cout << f(42) << std::endl;

    return 0;
}

I get the following output:

ctor
42
dtor

While I know this is correct, I'm not certain why. Is there anyone with stdc++ knowledge who can explain it to me?

回答1:

Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created. [12.2/3]

回答2:

12.2 Temporary objects, clause 3: "Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created."

1.9 Program execution, clause 12: "A full-expression is an expression that is not a subexpression of another expression."

回答3:

Since there is a constructor which can accept the argument passed into the F() function the complier creates the object on the fly before putting the arguments on the stack. As can be see in the disassembly below. literal numbers are treated by default as ints so there is an acceptable conversion.

001115C5  call        implicit_t::implicit_t (11112Ch) 
001115CA  mov         dword ptr [ebp-4],0 
001115D1  mov         esi,esp 
001115D3  mov         eax,dword ptr [__imp_std::endl (11A308h)] 
001115D8  push        eax  
001115D9  lea         ecx,[ebp-0D4h] 
001115DF  push        ecx  
001115E0  call        f (111113h) 

Your temp object hangs around until the expression is fully evaluated. this can be made more evident if you add another call to your function.

int main()
{
    std::cout << f(42) << std::endl;

    std::cout <<f(80) << std::endl;

    return 0;
}

Which has an output of

ctor
42
dtor
ctor
80
dtor

回答4:

f(42) constructs an unnamed implicit_t implicitly. It lives for the duration of it's containing scope, just as any auto variable would. Naturally, the d'tor gets called on return 0; of main().