c++11 default initialization / value initialization / direct initialization

Question

I am mainly confused by the notion of different initialization in c++, particularly in c++11 standard.

I am specifically confused by the case where the object is constructed through new (relevant or not), and a list of initializers (which matches a user-defined constructor) are supplied, and that constructor does not initialize all members (including PODs and other classes).

The following demonstrates my confusion.

class B
{
public:
    int b1;
    int b2;
};

class C
{
public:
    int c1;
    int c2;
    C()
    {
        c2 = 1234;
    }
};

class A
{
public:
    int a1;
    B a2;
    C a3;
    A(int a): a1(a){}
};

Now if I write auto aptr = new A(5), can we comment on

what will be the value of each fields in aptr->a1, aptr->a2, aptr->a3?
what initialization is triggered during this process?

Supplementary: A part of the confusion comes from the example included in https://en.cppreference.com/w/cpp/language/zero_initialization

struct A
{
    int i;
    A() { } // user-provided default ctor, does not initialize i
};

struct B { A a; }; // implicitly-defined default ctor

std::cout << B().a.i << '\n'; // value-initializes a B temporary
                              // leaves b.a.i uninitialized in C++03
                              // sets b.a.i to zero in C++11
// (note that B{}.a.i leaves b.a.i uninitialized in C++11, but for 
// a different reason: in post-DR1301 C++11, B{} is aggregate-initialization,
// which then value-initializes A, which has a user-provided ctor)

Where the value initialization of B seems to trigger A's certain initialization (which one is it) that sets A.i. Any comment on this one?

For me, I would just ALWAYS define constructors that initialize ALL members and then you don't have to worry about what the compiler may or may not do. — Jim Rhodes
– Jim Rhodes, Commented Oct 10, 2019 at 21:49
Thanks - I understand. But it would be good to know the actual rules. — user10150293
– user10150293, Commented Oct 10, 2019 at 22:33

JeJo · Accepted Answer · 2019-10-11 06:58:48Z

1

What will be the value of each fields in aptr->a1, aptr->a2, aptr->a3?

aptr->a1 will have 5 : because of A(int a): a1(a){} and you pass A(5).
aptr->a2 will be a default (compiler gnerated) constructed B object(i.e. aggregate type) having its b1 and b2 garbage value.
aptr->a3 will be a C object constructed by the default constructor provided by you (i.e. C(){ c2 = 1234; }). Meaning c1 in a3 will have garbage value, and the c2 will be initlized with 1234.

answered Oct 11, 2019 at 6:58

JeJo

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2 Comments

user10150293 Over a year ago

Thanks, would you explain the code example in the supplement of the question?

JeJo Over a year ago

@LiangLuo That itself makes me confusing and that is why I did not consider in my answer. ;) On the other hand, if you are using c++11 or latter the latter case seems to be something important in the given comments. If you want specific reasoning about the code example, I would suggest asking a different question focussing only on it.

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