Is object guaranteed to be initialized before calling its member function?

Question

from this question (still not solved) I come across this new problem,so here I made an example:

//main.cpp  

int main() {
    return 0;
}

//file1.cpp
#include "b.h"
B b;

//file2.cpp
#include "a.h"
A a;

//a.h
#pragma once
#include<iostream>
#include "b.h"
extern B b;
class A
{
public:
    A(){ std::cout << "a cotr" << std::endl;b.Use(); }
};

//b.h
#pragma once
#include<iostream>
class B
{
public:
    B() { std::cout << "b ctor"<<std::endl; };
    void Use() { std::cout << "use b" << std::endl; }
};

In g++ 6.3.0 the output is:( g++ -o test file1.cpp file2.cpp prac.cpp -std=c++11)
a cotr
use b
b ctor

So from the code example it seems that there's no such guarantee,and probably is an undefined behavior ?Is there anywhere the standard said about this?(I do not think it's a duplicate because this situation is a little different:in the initialization of a,invoke b's member function.)

You'd probably want to read [this](https://isocpp.org/wiki/faq/ctors#static-init-order) too — Passer By, Apr 13 '18 at 12:54
@PasserBy I read that,but this situation is a little different,because in the initialization of `a`,invoke `b`'s member function. — choxsword, Apr 13 '18 at 12:55
@bigxiao invoking member function does not affect object creation, it is already created or not. — Slava, Apr 13 '18 at 12:56
@Slava So is the `b.Use()` invoking **undefined behavior** ? — choxsword, Apr 13 '18 at 13:00
@bigxiao if constructor have not been called yet then yes it is UB. Order of initialization of global objects in different compilation units is implementation specific. — Slava, Apr 13 '18 at 13:02
I agree this is a duplicate. Calling a member function rather than using the value is neither here nor there. — Martin Bonner supports Monica, Apr 13 '18 at 13:05
@MartinBonner What if we access `b`'s member object in `Use()`, or in `a`'s constructor? — choxsword, Apr 13 '18 at 13:08
@bigxiao accessing a member has no effect on when an object is initialized. — eerorika, Apr 13 '18 at 13:09
@user2079303 I'm just responding to what he said about "using the value". — choxsword, Apr 13 '18 at 13:10
"using the value" was in the context of the other question, where the objects are primitive types. Calling a member function of `b` *uses* `b`, and is UB before `b` is constructed. — Martin Bonner supports Monica, Apr 13 '18 at 13:12

Slava · Accepted Answer · 2018-04-13T13:48:46.357

1

Is object guaranteed to be initialized before calling its member function?

No, it is your job not to call any non-static member function on invalid object, that includes but not limited to passing nullptr as this, object not created yet or already destroyed object. One of the solution of this situation is to have static local object in a function instead of global one and return reference/pointer to it. That method still have problem with destruction order, but at least half of the problem is gone.

edited Apr 13 '18 at 13:48

answered Apr 13 '18 at 13:00

Slava

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So you mean that calling non-static member function of `obj.foo()` before `obj` is construted is undefined behavior?What about static member fuction? – choxsword Apr 13 '18 at 13:39
Static member functions are fine (they are basically no different to a global function) as long as they don't access any static member variables which aren't yet initialised. – Alan Birtles Apr 13 '18 at 13:42

Is object guaranteed to be initialized before calling its member function?

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