Why does decltype not see the member declaration? [duplicate]

Answer 1

If C++14 is available, you can omit the trailing return type, thus avoiding referencing a member outside function body and also making the code more compact:

auto begin() {
    return identities.begin();
}

In general it's more correct to use decltype(auto) for such forwarding methods, so return type can be a reference, though in this particular case it's the same (thanks to @Nawaz):

decltype(auto) begin() {
    return identities.begin();
}

Answer 2

From N3337 [basic.lookup.unqual]/7:

A name used in the definition of a class X outside of a member function body or nested class definition shall be declared in one of the following ways:

• before its use in class X or be a member of a base class of X, or...

Because the trailing return type is part of the function declaration rather than the definition, it can't look ahead to see what else is declared in the class, so you need to declare that member above the function declaration.

Answer 3

Move the declaration of the member before the function like so:

#include <vector>
struct M
{
   // implementation
   private:
      std::vector<int> identities;
   public:    
   // interface
     auto begin() -> decltype(identities.begin())
     {
        return identities.begin();
     }


};