4 ms·

Type of x is a very explicit literal type. Type of y is also fully explitit, it is a 'return type of f()'. Not any different from a type of a sub-expression.

• fauigerzigerk 10y ago
  The type of f() is explicitly specified somewhere outside the local context. The type of x is not explicitly specified anywhere. It is inferred from the rhs expression.
  • sklogic 10y ago
    The type 'return type of f()' is still a type. It does not matter that it is not normalised, most of the C++ types are used in a non normal form. Type 'struct MyKewlMegaStructure' is not any different and not any more "local".
    • fauigerzigerk 10y ago
      How does that mean that the type of x is not inferred from something outside the local context?
      • sklogic 10y ago
        Ok. Then in 'struct _abc x;' we also have a type inference. And in any sub-expression there is also a "type inference". Although I think this kind of twisting the common term definitions is totally pointless. "Type inference" got a very well defined meaning, which got nothing to do with any kind of a type propagation - the latter term also existing for a reason, to designate a certain sort of type systems, fundamentally different from the inference-based ones.
        • fauigerzigerk 10y ago
          Then a lot of what people write on this subject seems to be using incorrect terminology, including tons of stuff issued by computer science departments and wikipedia. After digging a little deeper into the history of this terminology I have to concede that you and catnaroek are right. There was from the beginning in the 1950s a distinction that I didn't know about. So I was wrong. Thanks to you both for enlightening me.