decltype for class method type

Question

I would like to store return value of class member function in another class.

This seems to work:

class Foo
{
public: 
   Foo(int) {} //non default constructor that hides default constructor
   unspecified_return_type get_value();


};

class Bar
{
    // stores a value returned by Foo::get_value
    decltype(Foo().get_value()) value;
};

However there is a reference to default constructor of class Foo, which may not be defined in some cases. Is there any way to do it without explicitly referring to any constructor?

Answer 1

Yup, there is. std::declval was introduced for exactly this reason (not needing to rely on a particular constructor):

decltype(std::declval<Foo>().get_value()) value;

Answer 2

You could do it with the help of std::declval like the example below:

#include <iostream>
#include <utility>

struct test {
  int val = 10;
};

class Foo {
public:
   test get_value() { return test(); }
};

class Bar {
public:
  using type = decltype(std::declval<Foo>().get_value());
};

int main() {
  Bar::type v;
  std::cout << v.val << std::endl;
}

LIVE DEMO

std::declval<T> converts any type T to a reference type, making it possible to use member functions in decltype expressions without the need to go through constructors.

std::declval is commonly used in templates where acceptable template parameters may have no constructor in common, but have the same member function whose return type is needed.