We have popFrontN already in std.range. It's specialized for ranges with
slicing.
The problem is that the '$' operator is not implemented yet. When it
will, infinite ranges that want to offer slicing may define it to return
a particular symbolic value. Then you can use r[5 .. $] to skip the
first 5 elements of the range.
Andrei
On 08/14/2010 07:29 PM, David Simcha wrote:
I've thought some more about how to support slicing in the case of
take(someInfiniteRange, someNumber) and I think it would be a good idea
to require infinite random-access ranges to support an advance()
function. This would basically be like slicing, but only change the
beginning of the range, not the end (since there is no end of an
infinite range). Calling infiniteRange.advance(N) would be equivalent to
calling infiniteRange.popFront() N times, except that advance() would
have O(1) time complexity. Sound good? Any alternative suggestions?
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