> If it is not SomeType, then ::name is a typedef for int and the parse tree is ... | Hacker News

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		petters on Oct 4, 2019 \| parent \| context \| favorite \| on: Parsing C++ is literally undecidable (2013) > If it is not SomeType, then ::name is a typedef for int and the parse tree is declaring a pointer-to-int named x. Doesn't this require "typename" at the beginning of the line?

MauranKilom on Oct 4, 2019 [–]

No, not in this instance. typename is needed to mark dependent names as types, but you can only have dependent (i.e. on a template parameter) names within templates.

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