How do I declare a template of function with inout parameter in Cppfront?

[lauler1]

Hi,

I have a silly question on the syntax in cppfront.

I have a function like: myfunc: (inout var: Mytype) = { … }.

I would like to pass this function myfunc as a parameter to another function using std::function.

I tried anotherfunc: (funcpar: std::function<void(Mytype&)>) but Cppfront does not accept the &.
If I remove the & then c++ complains that the types don't match.

How do I declare a template of function that contains inout parameter in Cppfront?

Thanks

[JohelEGP]

While we lack support (which #526 aims to remedy),
you can use std::function<void(std::add_lvalue_reference_t<Mytype>)>.

[lauler1]

Worked!

Thank you very much!

[hsutter]

You can also std::function< decltype(f) > where f is a function with the intended signature.

This works:

Mytype: type = { }

myfunc: (inout var: Mytype) = { std::cout << "myfunc"; }

main: () = {
    f: std::function< decltype(myfunc) > = myfunc&;
    m: Mytype = ();
    f(m);
    _ = m;
}

[ntrel]

I tried anotherfunc: (funcpar: std::function<void(Mytype&)>) but Cppfront does not accept the &.

(cppfront does but rewrites it to void(&Mytype), causing a Cpp1 error). I'm a bit surprised cppfront even accepts void(Mytype) as a template argument, it doesn't allow it in a type alias.

Presumably at some point this will work:

myfunc: (inout var: Mytype);
anotherfunc: (funcpar: *(inout Mytype) -> void);

anotherfunc(myfunc&);

[JohelEGP]

I'm a bit surprised cppfront even accepts void(Mytype) as a template argument, it doesn't allow it in a type alias.

Cpp2 parses void(Mytype) as an expression.
Cpp1 interprets it as a type template argument.
See the second point of the second list: #628 (comment).