Re: ushort calls byte overload

[Oleg B via Digitalmars-d-learn]

On Tuesday, 30 May 2017 at 21:42:03 UTC, Daniel Kozak wrote:
Compiler do many assumptions (it is sometimes useful).

but if compiler find one-to-one correspondence it don't make 
assumptions, like here?

import std.stdio;

void f(ushort u)
{
    writeln("ushort");
}

void f(ubyte u)
{
    writeln("ubyte");
}

void main()
{
    ushort y = 0;
    immutable ushort x = 0;

    f(y);
    f(x);
}

RESULT IS:
ushort
ushort

or it can?