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authorRob Speer <rspeer+gh@luminoso.com>2017-01-19 02:51:29 -0500
committerRob Speer <rob@luminoso.com>2017-02-10 01:31:14 -0500
commit5cc5e0851e5bd14b9855462408ecb9058ed5eaad (patch)
tree2b73c3d644ff1b9b434ddf681a1d04d7c266c9a9 /src/test/run-pass/thinlto
parent24a70eb598a76edb0941f628a87946b40f2a1c83 (diff)
downloadrust-5cc5e0851e5bd14b9855462408ecb9058ed5eaad.tar.gz
rust-5cc5e0851e5bd14b9855462408ecb9058ed5eaad.zip
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.
Diffstat (limited to 'src/test/run-pass/thinlto')
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