Captureless lambda cannot be converted to function pointer when stored in std::function

Question

Usually, a C++ lambda without a capture should be convertable to a c-style function pointer. Somehow, converting it using std::function::target does not work (i.e. returns a nullptr), also the target_type does not match the signature type even though it seems to be the same.

Tested on VC13 and GCC 5.3 / 5.2.0 / 4.8

Minimal testing example:

#include <functional>
#include <iostream>

void Maybe() {

}

void callMe(std::function<void()> callback) {
    typedef void (*ftype)();
    std::cout << (callback.target_type() == typeid(ftype)) << std::endl;
    std::cout << callback.target<ftype>() << std::endl;
}

int main() {
    callMe([] () {});
    callMe(Maybe);
}

expected output would be

1
<address>
1
<address>

actual output

0
0
1
<address>

The question is: Why does the lambda's signature differ from the passed function?

"... should be convertable to a ..." - citation needed.

Useless
– Useless

2016-01-13 14:10:04 +00:00
Commented Jan 13, 2016 at 14:10 — Useless
– Useless, Commented Jan 13, 2016 at 14:10
@Useless Hmmm, citation is here.

πάντα ῥεῖ
– πάντα ῥεῖ

2016-01-13 14:11:27 +00:00
Commented Jan 13, 2016 at 14:11 — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Jan 13, 2016 at 14:11
@πάνταῥεῖ thanks, i included it in the question

John Watson
– John Watson

2016-01-13 14:13:36 +00:00
Commented Jan 13, 2016 at 14:13 — John Watson
– John Watson, Commented Jan 13, 2016 at 14:13

Quentin · Accepted Answer · 2016-08-16 07:52:27Z

18

In your first call, std::function does not bother with decaying the lambda into a pointer, it just stores it, with its actual type (which is indeed not void()).

You can force the lambda to decay into a pointer before constructing the std::function with the latter by simply using a unary +:

callMe(+[](){});
//     ^

edited Aug 16, 2016 at 7:52

answered Jan 13, 2016 at 14:14

Quentin

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4 Comments

Konrad Rudolph Over a year ago

+1, But: I really dislike this + hack for integral/pointer type coercion. It just obscures the intent too much. Better make it explicit by wrapping this into a dedicated function.

πάντα ῥεῖ Over a year ago

So we found a dupe of this Q&A?

Quentin Over a year ago

@πάνταῥεῖ They are not strictly duplicates (the other ones are about type deduction and overload resolution), but they are indeed quite strongly related.

Quentin Over a year ago

@KonradRudolph It's a matter of getting used to it, I guess. Hiding it behind, say, a decay(T&&) function wouldn't bother me either.

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