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   Hi,   
      
   I was wondering about the usage std::forward in the std::function   
   underlying invoke operation:   
   20.8.11.2.4 – operator(A… a) – INVOKE(f, std::forward(a)…, R)   
      
   Will this be equivalent to simply using - INVOKE(f,static_cast(a)…,   
   R)?   
   This will mainly save annoying step-into std::forward during   
   debugging.   
      
   For example std::function:   
      int operator(int a, int b)   
        { return invoke-me(forward(a), forward(b))   
   vs.   
      int operator(int a, int b)   
        { return invoke-me(static_cast(a), static_cast(b))   
      
   Here is my analysis of std::forward vs. static_cast in the   
   context of std::function invocation:   
      
   1) A = T& -> std::forward == std::forward returning T&&& hence   
   T& == static_cast = static_cast   
   2) A = T&& -> std::forward == std::forward returning T&&&&   
   hence T&& == static_cast = = static_cast   
   3) A = T -> std::forward == std::forward returning T&& ==   
   static_cast = static_cast   
      
   Altogether std::forward is equivalent to static_cast in the   
   above (I know that this is not true in general, e.g. make_shared).  Am   
   I missing something?   
      
   Thanks,   
   Rani   
      
      
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