fusilli_jerry

قبل 7 سنواتdiscuss

This is still not ambiguous: in order for 'foo::a' to be a template, 'a' must be prefixed with the 'template' disambiguator keyword, and in fact for any of a, b, or c to be types at all they would have to be prefixed with 'typename'. As written, it must parse as two expressions; if 'foo::a' turns out to be a class template at instantiation time, it's an error.

fusilli_jerry

قبل 7 سنواتdiscuss

To be clear here though, you can always (*undecidably, but subject to practical constraints) fully parse a template definition into a parse tree, and that parse tree will not change for any instantiation.

In your example, 'a' is known at parse time to either be a type or a value even if it's a template parameter, so the statement will always parse one way or another. Of course, 'a' itself may change in type or value so the problem is still unwieldy and, probably key to your use case, requires knowledge of type information from a possibly far-off part of the translation unit.

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