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debug deduced type? #36

@ehuhtala

Description

@ehuhtala

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);
}
}

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