Viewing profile — fusilli_jerry

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 a…

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 chang…