Hi, is it possible to get address of generic function instance for specified arguments without calling the function?

Example:

auto foo(alias fn, Args...)(auto ref Args args){
    ///return function/delegate type of `fn` for arguments `args`
}

void main(){

    long x;

    auto fn = foo!((a, b) => true)(new int(42), x);

    static assert(is(typeof(fn) == bool function(int*, scope ref long)@safe pure nothrow @nogc ));
}

Fix:

auto fn = foo!((a, ref b) => true)(new int(42), x);

On Wednesday, 27 April 2022 at 15:23:26 UTC, vit wrote:

auto foo(alias fn, Args...)(auto ref Args args){
    ///return function/delegate type of `fn` for arguments `args`
}

void main(){

    long x;

    auto fn = foo!((a, b) => true)(new int(42), x);

    static assert(is(typeof(fn) == bool function(int*, scope ref long)@safe pure nothrow @nogc ));
}

yeah sure; declare an alias that (fully) specialize the generic func and take the address using the alias Identifier.

On Wednesday, 27 April 2022 at 16:03:32 UTC, user1234 wrote:

auto foo(alias fn, Args...)(auto ref Args args){
    ///return function/delegate type of `fn` for arguments `args`
}

void main(){

    long x;

    auto fn = foo!((a, b) => true)(new int(42), x);

    static assert(is(typeof(fn) == bool function(int*, scope ref long)@safe pure nothrow @nogc ));
}

This work for types but not for attributes like scope, return and auto ref.

On Wednesday, 27 April 2022 at 17:22:14 UTC, vit wrote:

Oh sorry... auto ref... I totally forgot this old bug