Simple C++11 hash function won't compile - c++

Question

The following code for a simple hash function won't compile
#include <cstddef>
#include <functional>
namespace {
struct Foo {
long i;
};
}
namespace std {
template<> struct hash<::Foo> {
size_t operator()(::Foo foo) const {
return hash<decltype(foo.i)>(foo.i);
}
};
}
My 4.8.5 g++ compiler emits these messages:
$ g++ -std=c++11 a.cpp
a.cpp: In member function ‘std::size_t std::hash<{anonymous}::Foo>::operator()({anonymous}::Foo) const’:
a.cpp:13:47: error: no matching function for call to ‘std::hash<long int>::hash(long int&)’
return hash<decltype(foo.i)>(foo.i);
^
a.cpp:13:47: note: candidates are:
In file included from /usr/include/c++/4.8.2/bits/basic_string.h:3033:0,
from /usr/include/c++/4.8.2/string:52,
from /usr/include/c++/4.8.2/stdexcept:39,
from /usr/include/c++/4.8.2/array:38,
from /usr/include/c++/4.8.2/tuple:39,
from /usr/include/c++/4.8.2/functional:55,
from a.cpp:2:
/usr/include/c++/4.8.2/bits/functional_hash.h:107:3: note: constexpr std::hash<long int>::hash()
_Cxx_hashtable_define_trivial_hash(long)
^
/usr/include/c++/4.8.2/bits/functional_hash.h:107:3: note: candidate expects 0 arguments, 1 provided
/usr/include/c++/4.8.2/bits/functional_hash.h:107:3: note: constexpr std::hash<long int>::hash(const std::hash<long int>&)
/usr/include/c++/4.8.2/bits/functional_hash.h:107:3: note: no known conversion for argument 1 from ‘long int’ to ‘const std::hash<long int>&’
/usr/include/c++/4.8.2/bits/functional_hash.h:107:3: note: constexpr std::hash<long int>::hash(std::hash<long int>&&)
/usr/include/c++/4.8.2/bits/functional_hash.h:107:3: note: no known conversion for argument 1 from ‘long int’ to ‘std::hash<long int>&&’
$ fg
The problem appears to be the call-by-reference in the first error message but I don't understand why or how to fix it.

Answer 1

You are missing a set of parentheses in
return hash<decltype(foo.i)>(foo.i);
In the above you are trying to construct a std::hash, not call its operator(). You need
return hash<decltype(foo.i)>()(foo.i);
// or
return hash<decltype(foo.i)>{}(foo.i);
where the empty set of parentheses/curly braces constructs the hash object, and the second set calls its operator()