Update String Deref to explain why using &String does not always work
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@ -144,7 +144,7 @@ use boxed::Box;
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/// # Deref
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///
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/// `String`s implement [`Deref`]`<Target=str>`, and so inherit all of [`str`]'s
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/// methods. In addition, this means that you can pass a `String` to any
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/// methods. In addition, this means that you can pass a `String` to a
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/// function which takes a [`&str`] by using an ampersand (`&`):
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///
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/// ```
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@ -160,8 +160,31 @@ use boxed::Box;
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///
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/// This will create a [`&str`] from the `String` and pass it in. This
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/// conversion is very inexpensive, and so generally, functions will accept
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/// [`&str`]s as arguments unless they need a `String` for some specific reason.
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/// [`&str`]s as arguments unless they need a `String` for some specific
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/// reason.
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///
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/// In certain cases Rust doesn't have enough information to make this conversion,
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/// known as deref coercion. For example, in this case a string slice implements
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/// a trait and the function takes anything that implements the trait, Rust would
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/// need to make two implicit conversions which Rust doesn't know how to do. The
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/// following example will not compile for that reason.
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///
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/// ```compile_fail,E0277
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/// trait TraitExample {}
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///
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/// impl<'a> TraitExample for &'a str {}
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///
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/// fn example_func<A: TraitExample>(example_arg: A) {}
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///
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/// fn main() {
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/// let example_string = String::from("example_string");
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/// example_func(&example_string);
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/// }
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/// ```
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///
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/// What would work in this case is changing the line `example_func(&example_string);`
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/// to `example_func(example_string.to_str());`. This works because we're doing the
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/// conversion explicitly, rather than relying on the implicit conversion.
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///
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/// # Representation
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///
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