CODEWITHSUNDEEP
c++constructorinitialization
Klaus
Klaus

Reputation: 25663

object construction with assignment operator

I am wondering that object construction with assignment operator works, never seen that before I saw this question:

return by value calls copy ctor instead of move

Reduced example code:

class A
{
    public:
        int x;
        A(int _x):x(_x){ std::cout << "Init" << std::endl;}
        void Print() { std::cout << x << std::endl; }
        A& operator = ( const int ) = delete;
};

int main()
{
    A a=9;
    a.Print();
}

Is writing of

A a(9);
A a{9};
A a=9;

all the same?

Upvotes: 1

Views: 1304

Answers (1)

songyuanyao
songyuanyao

Reputation: 173034

This has nothing to do with assignment operator, it's initialization, more precisely copy initialization, which just uses the equals sign in the initializer.

when a named variable (automatic, static, or thread-local) of a non-reference type T is declared with the initializer consisting of an equals sign followed by an expression.

For A a = 9; the approriate constructor (i.e. A::A(int)) will be invoked to construct a. 1

A a(9); is direct initialization, A a{9}; is direct list initialization (since C++11), they all cause the A::A(int) to be invoked to construct the object for this case. 2

1 Before C++17 the appropriate move/copy constructor is still required conceptually. Even though it might be optimized out but still has to be accessible. Since C++17 this is not required again.

2 Note that there're still subtle differences among these initialization styles, they may lead to different effects in some specialized cases.

Upvotes: 3

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