Would it be possible to output what a deduced type looks like?
E.g. for lambda, I'm thinking something like:
auto f = f=4{};
using f_type = std::decay<decltype(f)>::type;
constexpr {
compiler.debug($f_type);
}
Or for templates:
template
void func() {
constexpr {
compiler.debug($T);
}
}
Would it be possible to output what a deduced type looks like?
E.g. for lambda, I'm thinking something like:
auto f = f=4{};
using f_type = std::decay<decltype(f)>::type;
constexpr {
compiler.debug($f_type);
}
Or for templates:
template
void func() {
constexpr {
compiler.debug($T);
}
}