struct initialization using curly brackets

Question

Saw some codes like below in a C++ project:

struct Foo{ 
    std::wstring x; 
    //blah
}

// this func returns a Foo object 

Foo getFoo(){ 
    //blah 
}

void main() { 
    Foo obj{getFoo()}; //why can initialize by another Foo object in {}? 
}

{} is list-initialization. But no Foo arguments are listed here. Why does this work? Does struct have default copy constructor?

And does Foo obj(getFoo()) work? Any difference from the way of using {}?

1. Foo has an implicitly declared copy constructor; 2. Foo obj(getFoo()) works too; 3. Same effect here. — songyuanyao

Dr. Watson Dr. Watson · Accepted Answer · 2018-09-13T01:27:00

This is copy initialization. It calls the implicitly declared copy-constructor. Sources: https://en.cppreference.com/w/cpp/language/copy_initialization, https://en.cppreference.com/w/cpp/language/copy_constructor

struct initialization using curly brackets

1 Answers