[phobos] advance() for infinite random access ranges?

[Andrei Alexandrescu]

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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