countUntil's constraints

[Nicholas Wilson]

On Wednesday, 8 August 2018 at 01:33:26 UTC, Steven Schveighoffer 
wrote:
> On 8/7/18 9:20 PM, Nicholas Wilson wrote:
>> the first overload is
>> 
>> ptrdiff_t countUntil(alias pred = "a == b", R, Rs...)(R 
>> haystack, Rs needles)
>> if (isForwardRange!R
>> && Rs.length > 0
>> && isForwardRange!(Rs[0]) == isInputRange!(Rs[0])
>> && is(typeof(startsWith!pred(haystack, needles[0])))
>> && (Rs.length == 1
>> || is(typeof(countUntil!pred(haystack, needles[1 .. $])))))
>> 
>> What does `isForwardRange!(Rs[0]) == isInputRange!(Rs[0]` mean 
>> here?
>> Is it just the same as `isForwardRange!(Rs[0])`? Why is it 
>> written like that?
>
> No, not exactly the same.
>
> Superficially, this rejects elements that are input ranges but 
> NOT forward ranges. Other than that, I can't tell you the 
> reason why it's that way.
>
> -Steve

Ahhh, Rs[0] is not necessarily a range, consider:

`assert(countUntil("hello world", 'r') == 8);`

so  that means `isForwardRange!(Rs[0]) == isInputRange!(Rs[0]` if 
Rs[0] is a range it must be a forward range.

The second overload looks as though it will never be a viable 
candidate
ptrdiff_t countUntil(alias pred = "a == b", R, N)(R haystack, N 
needle)
if (isInputRange!R &&
     is(typeof(binaryFun!pred(haystack.front, needle)) : bool))
{
     bool pred2(ElementType!R a) { return binaryFun!pred(a, 
needle); }
     return countUntil!pred2(haystack); // <---
}

because the marked line can't recurse be cause there is only one 
arg, so it tries to call the first overload, which fails due to  
Rs.length > 0.