CODEWITHSUNDEEP
c++constructorinitializer-list
Kerrek SB
Kerrek SB

Reputation: 477650

Initializing members out of order - is this OK?

From a comment on this answer:

Class members are initialized in their order of declaration. By this logic, the following constructor should invoke undefined behaviour:

struct Foo
{
   Bar a;
   Bar b;

   Foo(Bar c) : a(b = c) { }
};

Patently, we are assigning to b first before a has been initialized. Assigning to an uninitialized object should be UB. It's not surprising that the code "works" with Bar = int, but if I make Bar a heavy class with constructors, I see that b does indeed get initialized before a.

(For extra lunacy, we can even say Foo(Bar c, Bar d) : a(b = c), b(d) { }, still with no warning.)

Yet GCC 4.6.1 doesn't warn about this. Is this acceptable, well-defined behaviour, or is this strictly wrong?

Upvotes: 4

Views: 263

Answers (1)

Cubbi
Cubbi

Reputation: 47498

Given a Bar where non-initialized state actually matters to the assignment operator, I get warnings from GCC:

#include <iostream>
struct Bar {
    int n;
    Bar(int v) : n(v) {
            std::cout << "Bar " << n << " constructed\n";
    }
    Bar& operator=(const Bar& other) {
        std::cout << "Bar " << n << " assigned from " << other.n << "\n";
        n = other.n;
        return *this;
    }
};
struct Foo
{
   Bar a;
   Bar b;

   Foo(Bar c, Bar d) : a(b = c), b(d) { }
};

int main()
{
        Foo f(Bar(1), Bar(2));
}

test: https://ideone.com/VDZzG

test.cc: In function ‘int main()’:
test.cc:8:32: warning: ‘*((void*)(& f)+4).Bar::n’ is used uninitialized in this function [-Wuninitialized]
test.cc:23:13: note: ‘*((void*)(& f)+4).Bar::n’ was declared here

Other compilers I've tried don't seem to care, though...

Upvotes: 4

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