Default argument using curly braces initializer

问题

I have this snippet of code which seems to work well:

class foo{/* some member variables and functions*/};
void do_somthing(foo x={}){}
int main(){
  do_somthing();
}

I used to use void do_somthing(foo x=foo()){} to default the x argument but I see this way ={} in some book or online example(can not remember). Is it totally ok to use it? Is there any difference between the two methods?

回答1:

foo x=foo() is copy initialization,

Initializes an object from another object

and foo() is value initialization.

This is the initialization performed when a variable is constructed with an empty initializer.

foo x={} is aggregate initialization.

Initializes an aggregate from braced-init-list

If the number of initializer clauses is less than the number of members and bases (since C++17) or initializer list is completely empty, the remaining members and bases (since C++17) are initialized by their default initializers, if provided in the class definition, and otherwise (since C++14) by empty lists, which performs value-initialization.

So the result is the same in this case (both value-initialized).

And the effects of value initialization in this case are:

if T is a class type with a default constructor that is neither user-provided nor deleted (that is, it may be a class with an implicitly-defined or defaulted default constructor), the object is zero-initialized

Finally the effects of zero initialization in this case are:

If T is a scalar type, the object's initial value is the integral constant zero explicitly converted to T.

If T is an non-union class type, all base classes and non-static data members are zero-initialized, and all padding is initialized to zero bits. The constructors, if any, are ignored.

来源：https://stackoverflow.com/questions/35910619/default-argument-using-curly-braces-initializer