isAnagramOf

Return true if s1 is an Anagram of s2.

Equal arguments are not considered to be an anagrams of each other.

TODO Is there a faster way of calculating anagrams? TODO Allow const input TODO Move to Phobos std.algorithm.sorting. TODO Should requirement isInputRange be relaxed?

Note that implementations in http://rosettacode.org/wiki/Anagrams#D doesn't correctly handle multi-byte encoded characters in string and wstring.

isAnagramOf
(
R1
R2
)
(
R1 r1
,
R2 r2
)
if (
isInputRange!R1 &&
isInputRange!R2
&&
areEquable!(ElementType!R1, ElementType!R2)
)

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