On 12/14/15 8:50 PM, tcak wrote:
Hiding conditionals does not seem like to be solution. Here is my idea:
Show the function:
bool isSameLength(Range1, Range2)(Range1 r1, Range2 r2)
Then show the conditions separately:
if(
isInputRange!Range1 &&
isInputRange!Range2 &&
!isInfinite!Range1 &&
!isInfinite!Range2
)
Here are all the overloads of find (each with differing documentation):
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).
My take is that the user only needs to know that if he wants to find a
needle in a haystack, he does:
auto result = find(haystack, needle [, needle2, ...]);
Find will handle just about any call you can think of. How can the docs
just say this concisely?
-Steve
