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Clarify the English translation of ?Sized #33747

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May 21, 2016
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Clarify the English translation of ?Sized #33747

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Clarify the English translation of `?Sized`
postmodern May 20, 2016
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Keep line-width within 80 columns
postmodern May 20, 2016
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postmodern May 20, 2016
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postmodern May 21, 2016
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7 changes: 3 additions & 4 deletions src/doc/book/unsized-types.md
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Expand Up @@ -47,14 +47,13 @@ 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 syntax, `?Sized`:
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I personally think this is better (and more correct) as "bound" than "syntax". Maybe "trait bound" to clarify further?

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Yeah, bound makes more sense to me too.

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I used syntax as ?Trait is unique only to Sized. Plus it removes or loosens a bounded trait, not adds a negating a trait?

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It's different bound, but still a bound.

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Does ?Sized remove the Sized trait, or adds a "not Sized" trait in addition to the implicit Sized trait?

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Also the documentation refers to ?Sized as a special syntax.

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I would defer to the book over the actual Sized docs here; I haven't gotten to them yet. I would describe this as a special bound. While it is also technically special syntax, it's like, a sub-syntax: it's still a bound, but ?Sized is a special case within the trait bound syntax. if that makes sense.


```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`”, allowing us to match both constant size and unsized types.
All generic type parameters implicitly have the `Sized` bound, so `?Sized` can be used to opt-out of the implicit bound.
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Small nit, the line wrapping here is different to the rest of the file, it should probably be whatever the rest of the text is (looks like 80 chars?).