'auto' not allowed in function prototype w

Question

Using Clang 3.5, 3.6, or 3.7, with the flag std=c++1y the following code does not compile :

#include <iostream>
auto foo(auto bar) { return bar; }
int main() {
  std::cout << foo(5.0f) << std::endl;
}

The error given is :

error: 'auto' not allowed in function prototype

I do not have errors using g++ 4.9. Is this error produced because Clang has not yet implemented this functionnality yet or is it because I am not allowed to do that and GCC somehow permits it ?

Answer 1

You can use a template instead:

template<class A>
A foo(A bar) { return bar; }

Auto is only allowed when the Compiler can deduce the type from the context.

Answer 2

As we see from the ISO C++ discussion mailing: decltype(auto) parameters vs. perfect forwarding auto parameters of non-lambdas is part of concepts lite and therefore not in C++14:

clang is correct in the sense that we don't yet have auto parameters. Concepts lite may bring those, but C++14 doesn't have them.

If we use the -pedantic flag with gcc we receive the following warning:

warning: ISO C++ forbids use of 'auto' in parameter declaration [-Wpedantic]
  auto foo(auto bar) { return bar; }
           ^

So this looks like an extension.

As dyp pointed out, polymorphic lambdas did make it into C++14 and do allow auto parameters, an example taken from the paper:

// 'Identity' is a lambda that accepts an argument of any type and
// returns the value of its parameter.
auto Identity = [](auto a) { return a; };
int three = Identity(3);
char const* hello = Identity("hello");

Which is incidentally the same functionality you want to implement in your example.

Answer 3

Although your specific syntax did not make it to C++14, a similar option which did is:

static auto foo = [](auto bar) { return bar; };

which achieves basically the same thing.

Answer 4

The compiler cannot infer the type from the context.

What's wrong with doing

template<typename Y>
Y foo(Y bar){return bar;}

and must you pass bar by value?

In your case you can use the trailing return type syntax:

auto foo(auto bar) -> decltype(bar)