On 12/14/15 9:34 PM, Steven Schveighoffer wrote:
InputRange find(alias pred = "a == b", InputRange, Element)(InputRange
haystack, Element needle) if (isInputRange!InputRange &&
is(typeof(binaryFun!pred(haystack.front, needle)) : bool));
InputRange find(alias pred, InputRange)(InputRange haystack) if
(isInputRange!InputRange);
R1 find(alias pred = "a == b", R1, R2)(R1 haystack, R2 needle) if
(isForwardRange!R1 && isForwardRange!R2 &&
is(typeof(binaryFun!pred(haystack.front, needle.front)) : bool) &&
!isRandomAccessRange!R1);
Tuple!(Range, size_t) find(alias pred = "a == b", Range,
Ranges...)(Range haystack, Ranges needles) if (Ranges.length > 1 &&
is(typeof(startsWith!pred(haystack, needles))));
Range1 find(Range1, alias pred, Range2)(Range1 haystack,
BoyerMooreFinder!(pred, Range2) needle);
If you can decipher this into what find actually will accept, then you
have more patience than me. I think what I wanted (it's difficult to
remember) is whether I could pass in a subrange for find to search for
as the needle. It appears to reject cases where the haystack is a random
access range and the needle is a range (but it does work).
Heh, looking at the source, I see I was a victim of improper documentation.
There are 2 additional overloads of find that are NOT DOCUMENTED.
It's likely that the 2 overloads vary only by constraints, and that's
why they are not documented.
But as a user, again, I shouldn't have to care what the constraints are
specifically.
-Steve
