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fusilli_jerry
searching PlanetScale…
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fusilli_jerry
7y ago
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 pr
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by
fusilli_jerry
7y ago
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 k