Evaluating if ( std::function == function )?

Question

How can I evaluate given expression (function<void()> == functor).

Code example:

#include <functional>
using namespace std;

void Foo() { }

int main() {

    function<void()> a;
    // if (a == Foo) -> error
}

Edit: debugging details, and remove picture.

No images of errors or code, or otherwise plain text please! — πάντα ῥεῖ, Mar 04 '22 at 14:24
`std::function` is a type-erasing wrapper around any kind of callable. AFAIK there is no way to check what function it actually wraps. `operator==` can only check if the `std::function` has a callable in it or not, not which one it is. — super, Mar 04 '22 at 14:29
IMO this is [XY problem](http://mywiki.wooledge.org/XyProblem). Please explain what your code should do, then we can provide proper solution which will not require comparing `std::function`. — Marek R, Mar 04 '22 at 14:33
Possible duplicate of [Comparing std::functions for equality?](https://stackoverflow.com/questions/20833453/comparing-stdfunctions-for-equality) — Wyck, Mar 04 '22 at 14:45

Drew Dormann · Accepted Answer · 2022-03-06T23:54:29.147

6

std::function::target() will retrieve a pointer to the stored callable object. And that's what you're trying to compare.

You must specify the expected stored type, and target() will return nullptr if the type does not match.

(This means that if std::function holds a function pointer, target() will return a pointer-to-a-function-pointer.)

auto f = Foo; // This pointer existed in an earlier version of the question

if (*it->target<decltype(f)>() == f) { // Success

See it work in Compiler Explorer

Note that because functions decay to function pointers:

*it->target< decltype(Foo) >() == Foo

will not work, because std::function can not possibly be holding a copy of the function Foo. You would need to decay the function, such as:

*it->target< std::decay_t<decltype(Foo)> >() == Foo

or:

*it->target< decltype(&Foo) >() == Foo

edited Mar 06 '22 at 23:54

answered Mar 04 '22 at 14:35

Drew Dormann

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I suppose the expected type is `decltype(f)`. – MSalters Mar 04 '22 at 14:39
Spent a couple hours trying to figure this out. Thanks. – GråSkærm Mar 04 '22 at 14:40
@MSalters Very good idea. Thank you. – Drew Dormann Mar 04 '22 at 14:40
Maybe as additional note: If `auto f = Foo;` hadn't decayed the type already, it should probably be `std::decay_t`, e.g. if `f` was the name of the function instead. – user17732522 Mar 04 '22 at 14:45
1

@user17732522 it should be, yes. `decltype(Foo)` would fail in this call. – Drew Dormann Mar 04 '22 at 14:46
1

I would also maybe add that if you don't already know the target type for sure, you will also need to special case for `target()` returning a null pointer. (You mentioned that, but in the code it is not considered.) – user17732522 Mar 04 '22 at 14:48
`(*it->target() == Foo` also works. But not `(*it->target() == Foo`. Must be that decay behavior not decaying when it has become so taken for granted. – Eljay Mar 04 '22 at 15:03

Evaluating if ( std::function == function )?

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