debug deduced type?

[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);
}
}

[asutton]

Fixed in the injection-2 branch.

[ehuhtala]

Works very well for the first case, thanks :)

Can the constexpr {} block be inside class/function scopes? or just global scope? e.g. the template example still doesn't work due to:

blah.cpp:102:5: error: expression is not an integral constant expression
constexpr {
^
blah.cpp:102:5: note: non-literal type '(lambda at blah.cpp:102:5)' cannot be used in a constant expression

[ehuhtala]

Looks like the error is specific to templated functions/classes. If they are non-templated, the constexpr{} block works as I'm expecting.

[ehuhtala]

in template classes, the error looks like:
blah.cpp:15:5: error: expression is not an integral constant expression constexpr { ^ blah.cpp:16:9: note: non-literal type '<dependent type>' cannot be used in a constant expression compiler.debug($T); ^ note: in call to '__constexpr_decl()'