Why is int implicitly convertible to ulong?

[Jonathan M Davis]

On Friday, February 21, 2014 12:41:07 Hannes Steffenhagen wrote:
> The specific problem here was when working with std.json.
> 
> std.json distinguishes between UINTEGER and INTEGER, so I had
> code like
> 
> static if(is(T : ulong)) {
>      // must be UINTEGER
> } else static if(is(T : long)) {
>      // can be either INTEGER or UINTEGER
> }
> 
> 
> I've since found out about isSigned and isUnsigned, still it was
> mighty confusing for me that the first case was selected for
> signed types.

Well, it is using : - which means implicit conversion, which is always 
something that you should be careful with in generic code. Technically, 
something like

struct S
{
    ulong ul;
    alias ul this;
}

would match it as well, and depending on what the code inside the static if is 
like, it might work, but there's also a good chance that it wouldn't. Using 
is(T == ulong) if you want to require that it be exactly ulong, or 
isUnsigned!T if you want any unsigned integral type. Using implicit conversion 
in generic code _can_ be useful, but you need to be _very_ careful with it.

- Jonathan M Davis