4 ms·

I think that breaks down as soon as templates are involved, which is very pervasive in C++ codebase. Given template<typename T, typename sometype> voi

by wbkang 8y ago

I think that breaks down as soon as templates are involved, which is very pervasive in C++ codebase. Given template<typename T, typename sometype> void foo(T<sometype>& baz) It's not so straightforward anymore.

jgtrosh 8y ago
It can still apply the operation anytime a specialized object is defined. (This is how templates are compiled, right?)