Thread overview
August 08
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?
August 07
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
August 08
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

But forward ranges are a superset of input ranges so `isForwardRange!(Rs[0]) == isInputRange!(Rs[0]` <=> `isForwardRange!(Rs[0])`, right?
August 08
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.

August 08
On 8/7/18 10:28 PM, Nicholas Wilson wrote:
> 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.
> 

Ah, but there is a third overload which just takes a haystack and a predicate.

What is a bit more confusing to me is why there isn't an ambiguity error when the needle parameter is one element that is not a range.

-Steve