Fix a misleading statement in Iterator.nth()

The `Iterator.nth()` documentation says "Note that all preceding elements will be consumed". I assumed from that that the preceding elements would be the *only* ones that were consumed, but in fact the returned element is consumed as well.

The way I read the documentation, I assumed that `nth(0)` would not discard anything (as there are 0 preceding elements), so I added a sentence clarifying that it does. I also rephrased it to avoid the stunted "i.e." phrasing.
This commit is contained in:
Rob Speer 2017-01-19 02:51:29 -05:00 committed by Rob Speer
parent 24a70eb598
commit 5cc5e0851e

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@ -209,7 +209,10 @@ pub trait Iterator {
/// Returns the `n`th element of the iterator.
///
/// Note that all preceding elements will be consumed (i.e. discarded).
/// Note that all preceding elements, as well as the returned element, will be
/// consumed. That means that the preceding elements will be discarded, and also
/// that calling `nth(0)` multiple times on the same iterator will return different
/// objects.
///
/// Like most indexing operations, the count starts from zero, so `nth(0)`
/// returns the first value, `nth(1)` the second, and so on.