Template syntax explanation

Question

On some resource I saw a code:

struct C
{
    template <typename T, typename = typename enable_if<!is_same<C, typename decay<T>::type>{}>::type>
    C(T&& f){}
}

I tried my best but still need some clarification. I know this scary thing is needed for the SFINAE idiom - if something fails the templated function simply will not be created. Here what I found:

1. typename decay<T>::type - this removes cv qualifiers from type T , or converts an array T to pointer T or converts T to function pointer. But what is this typename before? I suppose this is related to dependent type, ie the supplied type T is a thing that is a part of another template, right?
2. is_same<A, B>{} - what is this {} braces doing here? Why?
3. typename enable_if<A>::type - as I understood if A is true the type field exist and in this case it's void since only one argument was passed to enable_if , right? But again - what is this typename before?
4. template <typename T, typename = typename A> - what is this typename = typename A ? Where's the argument name at all?!

Answer 1

1. Yes, typename keyword in that case says to compiler that decay::type is really type (but not static variable of std::decay<T> , for example)
2. std::is_same has cast operator to bool, and through {} object of that type is created and then converted to bool (at compile time). Alternative variant is std::is_same<...>::value
3. If std::is_same returns true, then member type of std::enable_if is exists and SFINAE can pass through that function
4. Parameter name is not needed in that context (because it isn't used somewhere)

PS std::enable_if looks like that

template<bool B, class T = void>  //(1)
struct enable_if {};

template<class T>  /(2)
struct enable_if<true, T> { typedef T type; };

So, if the first parameter is false then primary template (1) is selected and it has not type member, but in case when it is true specialization (2) is selected