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
