How do I typedef a function pointer with the C++11 using syntax?

It has a similar syntax, except you remove the identifier from the pointer:

using FunctionPtr = void (*)();

Here is an Example

If you want to "take away the uglyness", try what Xeo suggested:

#include <type_traits>

using FunctionPtr = std::add_pointer<void()>::type;

And here is another demo.

The "ugliness" can also be taken away if you avoid typedef-ing a pointer:

void f() {}
using Function_t = void();    
Function_t* ptr = f;
ptr();

http://ideone.com/e1XuYc

You want a type-id, which is essentially exactly the same as a declaration except you delete the declarator-id. The declarator-id is usually an identifier, and the name you are declaring in the equivilant declaration.

For example:

int x

The declarator-id is x so just remove it:

int

Likewise:

int x[10]

Remove the x:

int[10]

For your example:

void (*FunctionPtr)()

Here the declarator-id is FunctionPtr. so just remove it to get the type-id:

void (*)()

This works because given a type-id you can always determine uniquely where the identifier would go to create a declaration. From 8.1.1 in the standard:

It is possible to identify uniquely the location in the [type-id] where the identifier would appear if the construction were a [declaration]. The named type is then the same as the type of the hypothetical identifier.