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Clarify the English translation of ?Sized #33747

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merged 4 commits into from
May 21, 2016
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Clarify the English translation of ?Sized #33747

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2fd4e60
Clarify the English translation of `?Sized`
postmodern May 20, 2016
d021d7d
Keep line-width within 80 columns
postmodern May 20, 2016
d8c086b
Grammar change
postmodern May 20, 2016
71af58a
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postmodern May 21, 2016
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9 changes: 5 additions & 4 deletions src/doc/book/unsized-types.md
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Expand Up @@ -47,14 +47,15 @@ pointers, can use this `impl`.
# ?Sized

If you want to write a function that accepts a dynamically sized type, you
can use the special bound, `?Sized`:
can use the special bound syntax, `?Sized`:

```rust
struct Foo<T: ?Sized> {
f: T,
}
```

This `?`, read as “T may be `Sized`”, means that this bound is special: it
lets us match more kinds, not less. It’s almost like every `T` implicitly has
`T: Sized`, and the `?` undoes this default.
This `?Sized`, read as “T may or may not be `Sized`”, which allows us to match
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Sorry, I think this is still not quite right...
My understanding is that you should be able to remove the appositive (the part between the commas) and still have a valid sentence. So I think the word "which" here is extraneous.

I'd prefer if someone else could chime in here as well on the grammar.

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I'm horrible with grammar, so I'll defer.

both sized and unsized types. All generic type parameters implicitly
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"which" should be removed from the first sentence.

have the `Sized` bound, so the `?Sized` can be used to opt-out of the implicit
bound.