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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 M
by 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 agoHow does that mean that the type of x is not inferred from something outside the local context?
- sklogic 10y agoOk. 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 agoThen 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.