Why does type deduction fail on (non-pointer-to) function types

Question

Starting with some metaprogramming code:

template<class... Ts>
class list {}; //a generic container for a list of types

template<class in_list_type>
class front //get the type of the first template parameter
{
    template<template<class...> class in_list_less_template_type, class front_type, class... rest_types>
    static front_type deduce_type(in_list_less_template_type<front_type, rest_types...>*);
public:
    typedef decltype(deduce_type((in_list_type*)nullptr)) type;
};

This code works fine for this:

typedef typename front<list<int, float, char>>::type type; //type is int

But fails to compile when the first item is a function type:

// no matching function for call to 'deduce_type'
typedef typename front<list<void (), float, char>>::type type;

I only have access to XCode at the moment and can't confirm whether this is simply an XCode bug. I'm using XCode 4.5.1, using Apple LLVM compiler 4.1.

Answer 1

When the template arguments to deduce_type are being deduced, front_type has void() as a candidate. However this would make deduce_type have type void ()() (function returning a function -- alias<void()>() if you assume template<typename T> using alias = T; to be in scope). This is an error and type deduction fails.

A solution is to have deduce_type return something like identity<front_type> , and type be an alias to typename decltype(deduce_type((in_list_type*)nullptr))::type .