Hello.

1st Novice question:

i want function, operates sometimes with char[], sometimes with ubyte[].
internally it works with ubyte.
i can use overloading:

void myFunc(ubyte[] arg) {...};
void myFunc(char[] arg) { ubyte[] arg2 = cast(ubyte[]) arg; ...}

It is OK. But i want 2 params (arg1, arg2),
so i need write 4 overloading functions.

I fill templated needed, can anybody show me the way?


And 2nd question:

what if additionally to written above, function shuld return ubyte[] or char[] ?
can compiler guess what return type need from code?
something like:

char[] cc = myFunc()
ubyte[] bb = myFunc()

Thanks.

On Sunday, 10 January 2016 at 10:10:46 UTC, zabruk70 wrote:
> Hello.
>
> 1st Novice question:
>
> i want function, operates sometimes with char[], sometimes with ubyte[].
> internally it works with ubyte.
> i can use overloading:
>
> void myFunc(ubyte[] arg) {...};
> void myFunc(char[] arg) { ubyte[] arg2 = cast(ubyte[]) arg; ...}
>
> It is OK. But i want 2 params (arg1, arg2),
> so i need write 4 overloading functions.
>
> I fill templated needed, can anybody show me the way?

No you actually don't have to write all possible overloads.

For example:
(surely not the best solution)

import std.traits;

ubyte[] myFuncImpl(ubyte[] a, ubyte[] b)
{
	return a ~ b;
}

ubyte[] myFunc(T1,T2)(T1[] a, T2[] b)
	if(is(Unqual!T1 == ubyte) || is(Unqual!T1 == char)
		|| is(Unqual!T2 == ubyte) || is(Unqual!T2 == char))
{
	return myFuncImpl(cast(ubyte[]) a, cast(ubyte[]) b);
}


> And 2nd question:
>
> what if additionally to written above, function shuld return ubyte[] or char[] ?
> can compiler guess what return type need from code?

No. Your function call signature would be completely the same..

The way to do this is to make the function a template and pass the return parameter to it.
So you're able to do something like:

auto val = myReturnFunc!int();


togrue

On Sunday, 10 January 2016 at 11:13:00 UTC, Tobi G. wrote:
> ubyte[] myFunc(T1,T2)(T1[] a, T2[] b)

Tobi, big thanks!!!
I should learn templates...

On Sunday, 10 January 2016 at 10:10:46 UTC, zabruk70 wrote:
> void myFunc(char[] arg) { ubyte[] arg2 = cast(ubyte[]) arg; ...}


void myFunc(const(void)[] arg) {
   const(ubyte)[] arg2 = cast(const(ubyte)[]) arg;
   // use arg2
}


A `const(void)[]` type can accept any array as input. void[] is any array, const means it will take immutable, const, and mutable as well. Then you cast it to one type to use it.

> what if additionally to written above, function shuld return ubyte[] or char[] ?

That depends on what you're doing, but you might also just want to return const(void)[], which the user will have to cast to something for them to use.

On Sunday, 10 January 2016 at 14:17:28 UTC, Adam D. Ruppe wrote:
> A `const(void)[]` type can accept any array as input. void[] is

Ah, how i can forget about void[] !
Thanks Adam!