AMDG Ovanes Markarian wrote:
Hicham,
it does not create a new new type it only declares a type.
Consider the following example:
template<int ()> struct X {};
This means your type X expects a function pointer type with return value int and no parameters. This can be used in a meta programming to make some type assumptions or inspection. This is how the result_of works. Think of result_of as of type X but a bit more complex and which can give an assumption about the function type passed. We can write:
template<int ()> struct X { typedef int type; };
<snip>
Here MyFunc is a name of the function pointer type, which allows us to reference this type from inside the template.
Hope that helps.
This has very little to do with result_of. The expression result_of<floatfct(float, float)>::type is the return type of calling an object of type floatfct with two float arguments. floatfct(float, float) is a function type that takes two floats and returns a floatfct. result_of is an ordinary class template, that uses metaprogramming to decompose the function type and treats the return type as the function that will be called. template<class T> struct result_of; template<class F> struct result_of<F()> {}; template<class F, class Arg1> struct result_of<F(Arg1)> {}; ... In Christ, Steven Watanabe