Re: Initializers

Victor Bazarov <>
Mon, 19 Apr 2010 10:54:38 -0400
Michael Tsang wrote:

I'm very confused among different types of initializations:

    scalar: set to 0
    class/array: initialize recursively
    union: initialize first member
    reference: no-op

Huh? Where do you get the "no-op"?

    class: call default constructor
    array: initialize recursively
    reference: ill-formed
    others: no-op

    class with user-provided constructor: call default constructor
    non-union class without user-provided constructor: zero-
initialise and call non-trivial constructor
    array: initialize recursively
    reference: ill-formed
    others: zero-initialize

implicitly-defined default constructor:
    default-initialize base classes and members recursively

When the direct initializer is (), the object is value-initialized but
consider the following:

#include <iostream>
struct A
        std::cout << "test" << std::endl;
    int x;

struct B
    // non-trivial implicit default constructor
    A a;
    int x;

A x; // x is zero-initialized and then default initialized so that x.x
== 0

int main()
    A *p = new A; // p->x is not determined.
    A *q = new A(); // q->x is also not determined?! A
standard caveat?!

Why is that a caveat? If you provide the default constructor and
*purposefully* omit 'x' from the initializer list, it's left
uninitialized no matter whether you use parens here or not.

     delete p;
    delete q;
    B *r = new B; // default-initialized recursively: s->x
and s->a.x are both indeterminate.
    B *s = new B(); // zero-initialized first: s->x and s->a.x
are both 0

Zero-initialized first? Where do you get that? It's not static.

The dynamic B object is value-initialized, yes (see 5.3.4/15)? And
since 'B' is an aggregate (no user-defined c-tors, no private or
protected non-static members, no virtual functions, no base classes), to
value-initialize it means that every non-static member and every base
class is value-initialized. And since 'A' has a user-defined c-tor, to
value-initialize it means to call that c-tor.

So, nope. Since 'A' has a user-declared c-tor, s->a.x is still


Also, how does zero-initializing a reference make sense?

There ain't no such a thing as "zero-initializing a reference".

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