CODEWITHSUNDEEP
c++initializationaggregate-initialization
Destructor
Destructor

Reputation: 14458

Is this aggregate initialization or default initialization in C++?

Consider following program.

#include <iostream>
int main()
{
    int a=int{};
    std::cout<<a;
}

Is it uses aggregate initialization or default initialization? I am confused.

Upvotes: 1

Views: 178

Answers (3)

JtTest
JtTest

Reputation: 187

Since C++11, by comparison with other SO answers (e.g.: this or this), I would say this is:

  1. a value-initialization (int{}) followed by
  2. a copy-initialization (int a=int{}).

By the way, from C++17, the second step should vanish as int{} is required to directly initialize a.

Upvotes: 0

eerorika
eerorika

Reputation: 238491

Empty parentheses or braces (T() or T{}) perform value initialization. The exception would be the case where the type is an aggregate in which case aggregate initialization would be used. Since int is not an aggregate, it will be value initialized and since it's not a class nor an array, value initialization will do zero-initialization.

You were wondering why it doesn't work in C. Such syntax simply doesn't exist in C, see this answer.

Upvotes: 5

Olivier Poulin
Olivier Poulin

Reputation: 1796

Aggregate initialization is a type of list initialization, which initializes aggregates. An aggregate is an object of type array, or object which has the characteristics defined on this page.

In this case, the type of initialization is most likely value initialization.

Upvotes: 2

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