c++ initializer difference, with and without = operator

Question

What is difference between initializing x1 and x2?

struct X {
    int i;
};

void func(){
    X x1 = {2};
    X x2 {2};

    cout << x1.i << ", " << x2.i << endl;
}

http://stackoverflow.com/questions/1051379/is-there-a-difference-in-c-between-copy-initialization-and-direct-initializati — 101010, Apr 03 '17 at 08:56
Possible duplicate of [Is there a difference in C++ between copy initialization and direct initialization?](http://stackoverflow.com/questions/1051379/is-there-a-difference-in-c-between-copy-initialization-and-direct-initializati) — Manoj, Apr 03 '17 at 09:02
@Manoj in that question, () operators used. I have used {} operator. — nullptr, Apr 03 '17 at 09:09

score 2 · Accepted Answer · answered Apr 03 '17 at 09:50

X x1 = {2};

Is copy-list-initialization.

X x2 {2};

Is direct-initialization.

Both syntaxes perform aggregate initialization, because X is an aggregate.

What is difference

Braces can not be elided with the direct-initialization form (until C++14).

In general, copy initialization only considers non-explicit constructors and conversion functions and direct initialization considers also explicit ones. However, this does not apply to aggregate initialization.

with and without = operator

The equals (=) character here isn't actually an operator. It is part of the syntax of copy initialization.

@UDPLover there is no performance difference between aggregate initialization and aggregate initialization. — eerorika, Apr 03 '17 at 10:31

c++ initializer difference, with and without = operator

1 Answers1