Guide: Patterns: use non-x variables in match blocks
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@ -3801,7 +3801,7 @@ the value to a name with `@`:
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let x = 1i;
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match x {
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x @ 1 ... 5 => println!("got {}", x),
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e @ 1 ... 5 => println!("got a range element {}", e),
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_ => println!("anything"),
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}
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```
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@ -3834,7 +3834,7 @@ enum OptionalInt {
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let x = Value(5i);
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match x {
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Value(x) if x > 5 => println!("Got an int bigger than five!"),
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Value(i) if i > 5 => println!("Got an int bigger than five!"),
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Value(..) => println!("Got an int!"),
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Missing => println!("No such luck."),
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}
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@ -3847,12 +3847,12 @@ with. First, `&`:
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let x = &5i;
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match x {
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&x => println!("Got a value: {}", x),
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&val => println!("Got a value: {}", val),
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}
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```
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Here, the `x` inside the `match` has type `int`. In other words, the left hand
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side of the pattern destructures the value. If we have `&5i`, then in `&x`, `x`
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Here, the `val` inside the `match` has type `int`. In other words, the left hand
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side of the pattern destructures the value. If we have `&5i`, then in `&val`, `val`
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would be `5i`.
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If you want to get a reference, use the `ref` keyword:
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@ -3861,11 +3861,11 @@ If you want to get a reference, use the `ref` keyword:
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let x = 5i;
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match x {
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ref x => println!("Got a reference to {}", x),
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ref r => println!("Got a reference to {}", r),
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}
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```
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Here, the `x` inside the `match` has the type `&int`. In other words, the `ref`
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Here, the `r` inside the `match` has the type `&int`. In other words, the `ref`
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keyword _creates_ a reference, for use in the pattern. If you need a mutable
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reference, `ref mut` will work in the same way:
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@ -3873,7 +3873,7 @@ reference, `ref mut` will work in the same way:
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let mut x = 5i;
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match x {
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ref mut x => println!("Got a mutable reference to {}", x),
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ref mut mr => println!("Got a mutable reference to {}", mr),
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}
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```
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