[BUG] Mutable anonymous functions cannot be passed as in arguments when capture variable

[filipsajdak]

In the current code (faacec05d8a752f68abf7354e64dbc72cbf11989), all anonymous functions are mutable.

Unfortunately, it broke passing anonymous functions with capture as in arguments. That means that the following code will not work:

fun: (f) -> _ = {
    return f();
}

main: () = {
    i := 42;
    std::cout << fun(:() i$) << std::endl;
}

Which generates (skipping boilerplate):

[[nodiscard]] auto fun(auto const& f) -> auto{
    return f(); 
}

auto main() -> int{
    auto i {42}; 
    std::cout << fun([_0 = std::move(i)]() mutable -> auto { return _0;  }) << std::endl;
}

and fail during compilation with the following errors:

bug_mutable_lambda.cpp2:2:12: error: no matching function for call to object of type 'const (lambda at bug_mutable_lambda.cpp2:7:22)'
    return f(); 
           ^
bug_mutable_lambda.cpp2:7:18: note: in instantiation of function template specialization 'fun<(lambda at bug_mutable_lambda.cpp2:7:22)>' requested here
    std::cout << fun([_0 = std::move(i)]() mutable -> auto { return _0;  }) << std::endl;
                 ^
bug_mutable_lambda.cpp2:7:22: note: candidate function not viable: 'this' argument has type 'const (lambda at bug_mutable_lambda.cpp2:7:22)', but method is not marked const
    std::cout << fun([_0 = std::move(i)]() mutable -> auto { return _0;  }) << std::endl;
                     ^
1 error generated.

Also, I now have an issue using the boost::ut library, as it stores lambdas (that often have capture lists) as const references.

Maybe we can prefix mutable lambda with the inout keyword and make the unnamed function in by default?

[JohelEGP]

See commit b9e73e06f93c9238b3d67a539d8276aa8b9db814. Your function expression needs to be changed from :() i$ to :() -> _ == i$.

[hsutter]

Thanks! I think this has been answered so I'll close it, but if I missed something please reopen.

[MaxSagebaum]

I ran into this yesterday. Took me quite a while to figure out what was wrong.

Intuitively I would say that using the constexpr marker == affects only the result and not the declaration of the lambda. But I do not have a better idea how to handle this. What was the original problem that required mutable for the lambdas?

[JohelEGP]

It's a fix or change in the conceptual model of what captures are. See the thread starting at https://github.com/hsutter/cppfront/commit/4bd0c0438f2d3fa65d3e65a55b17c3a296bd8bc3#commitcomment-134056070:

my conceptual model of captures is actually closer to them being additional local variables that are "static" (persistent across calls), and local variables are non-const by default.

that == also gives the "not mutable" syntax

[MaxSagebaum]

Thank you for the reference. Having muteable as the default for lambdas needs to be getting used to when coming from cpp where it is the other way round.