How to translate this macro which return a function pointer to Adept?

[ghost]

#define DW_SIGNAL_FUNC(a) ((void *)a)

I omitted this macro from dw.adept because I think it's not really needed and I can replace it with func &a. But when translating dwtest.c to Adept, I found it uses the macro intensively. This means I have to translate this macro to Adept, too. This is simply text-substitution in C, but in Adept, I really don't know how to do it as Adept doesn't support text-substitution.

[IsaacShelton]

func DW_SIGNAL_FUNC(a $T) ptr = a as ptr would be the equivalent

[ghost]

This seems not working. Please review.

[IsaacShelton]

What specifically isn't working about it? I tested it and it seems to work

import basics

func DW_SIGNAL_FUNC(a $T) ptr = a as ptr

func main {
    print(DW_SIGNAL_FUNC(256))
}

Can you provide the code that uses it where it doesn't work correctly? Thank you

[ghost]

Please search for DW_SIGNAL_FUNC on dw.h and dwtest.c and you will find the usage of it. a is the name of a function. DW_SIGNAL_FUNC returns a function pointer to the function named a. Your code will only work if a is a normal variable.

[IsaacShelton]

Oh, it might be because you have to get the address of the function you want to use, so for example:

c code:

HPRINT print = dw_print_new("DWTest Job", 0, 2, DW_SIGNAL_FUNC(draw_page), NULL);

would become

adept code:

print HPRINT = dw_print_new('DWTest Job', 0, 2, DW_SIGNAL_FUNC(func &draw_page), NULL);

you need the func & otherwise it won't know that you want the function instead of a variable (which doesn't exist).

Maybe this is why?