Predicate the given ranges are sorted by.
A range of the difference of r1 and r2.
In the case of multisets, considering that element a appears x times in r1 and y times and r2, the number of occurences of a in the resulting range is going to be x-y if x > y or 0 otherwise.
import std.algorithm.comparison : equal; import std.range.primitives : isForwardRange; //sets int[] a = [ 1, 2, 4, 5, 7, 9 ]; int[] b = [ 0, 1, 2, 4, 7, 8 ]; assert(equal(setDifference(a, b), [5, 9])); static assert(isForwardRange!(typeof(setDifference(a, b)))); // multisets int[] x = [1, 1, 1, 2, 3]; int[] y = [1, 1, 2, 4, 5]; auto r = setDifference(x, y); assert(equal(r, [1, 3])); assert(setDifference(r, x).empty);
Lazily computes the difference of r1 and r2. The two ranges are assumed to be sorted by less. The element types of the two ranges must have a common type.