Unpacking arguments of a functional parameter to a C++ template class -

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I have functional template arguments for template classes in C ++.

I would like to take a template class foo to get a template parameter funny 

  template & lt; Typename Fun & gt; Struct Foo {...}; Functions such as   
 such as  
  zero times (std :: string a, float b, char c) given {...}  
 then  foo < Bar & gt; :: args_t  will be equal to a typedef  
  std :: tuple & lt; Std :: string, float, four & gt;   
 Is this possible? (Use of  std :: tuple  is only for concrete. More generally, I am thinking that it is possible to do the pattern-matching on the arguments of a working template parameter.)  
 Point  
  template fu & lt; The typename A, typename B, typename c, typename, D like to avoid defining  foo  (* funny) (a, bb, cc) & gt; Structure foo {typedef std :: tuple & lt; A, B, C & gt; Args_t; };   
 for which both require fixed fixes of the function, and the function's argument and return type are clearly required to be provided as a template parameter (Variadic Template Defining  Foo  might possibly solve a problem, but what about the latter?)  
 Thanks!   
 
 
 Announce the primary template and do not execute it.  
  template & lt; Typename T & gt; Struct foo; // Unchanged Primary Template   
 Then provide a partial specialization that matches the function type as a template argument.  
  template & lt; Typename result, typename ... Args & gt; Straight foo & lieutenant; Results (ARG ...) & gt; {Args_t = std :: Google & lt; ARG ... & gt; };   
 As a nested type  
  foo < Decltype (bar) & gt; :: args_t

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