c++ - Why decltype isn't implicit? -

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Why type of expression is expected, why decltype can not be added to expression? 

  template & lt; Class X, Square Y, Class Z & gt; Auto Foo (X X, Y, Z Z) {std :: vector & lt; Decltype (x + y * z) & gt; Values; // valid in c ++ 11 / c ++ 14 // std :: vector & lt; X + y * z & gt; Values; // Invalid values. Push_back (x + y * z); Return value; Type Type Type - OK}

typed with Expression

C ++ 14 compilers will reduce the function return type based on return expressions. Why can not it be extended in any expression -> type 'conversion?

The same applies, why do I have to write: 

  std :: vector & lt; Decltype (declval  () + declval>  () * declval & lt; z & gt; ()) & gt; Values; Instead of

: 

  std :: vector & lt; X + Y * Z & gt; Values;   
   
 
 
 If the underlying permissions of  decltype  Some, 
  
 consider the following example: 
 
  struct tp {template  
  Output:   
 
 value parameter  
 type parameter  
 type Parameter   
 If  decltype  is passed to  t1.foo & lt; X * y + z & gt; (); , the wrong version of this is called  foo () . 
 
     
  •  C ++ policy to express what you are doing, and IMHO is a very good thing by the compiler when possible. 
    •  
  •  After this,  decltype  is only 8 characters 
    •  
 
.

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