216 lines
6.1 KiB
Plaintext
216 lines
6.1 KiB
Plaintext
error: this pattern takes a reference on something that is being dereferenced
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--> tests/ui/needless_borrowed_ref.rs:30:34
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LL | let _ = v.iter_mut().filter(|&ref a| a.is_empty());
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| ^^^^^^
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= note: `-D clippy::needless-borrowed-reference` implied by `-D warnings`
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= help: to override `-D warnings` add `#[allow(clippy::needless_borrowed_reference)]`
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help: try removing the `&ref` part
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LL - let _ = v.iter_mut().filter(|&ref a| a.is_empty());
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LL + let _ = v.iter_mut().filter(|a| a.is_empty());
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error: this pattern takes a reference on something that is being dereferenced
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--> tests/ui/needless_borrowed_ref.rs:34:17
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LL | if let Some(&ref v) = thingy {}
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| ^^^^^^
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help: try removing the `&ref` part
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LL - if let Some(&ref v) = thingy {}
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LL + if let Some(v) = thingy {}
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error: this pattern takes a reference on something that is being dereferenced
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--> tests/ui/needless_borrowed_ref.rs:36:14
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LL | if let &[&ref a, ref b] = slice_of_refs {}
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| ^^^^^^
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help: try removing the `&ref` part
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LL - if let &[&ref a, ref b] = slice_of_refs {}
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LL + if let &[a, ref b] = slice_of_refs {}
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error: dereferencing a slice pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:38:9
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LL | let &[ref a, ..] = &array;
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| ^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - let &[ref a, ..] = &array;
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LL + let [a, ..] = &array;
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error: dereferencing a slice pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:39:9
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LL | let &[ref a, ref b, ..] = &array;
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| ^^^^^^^^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - let &[ref a, ref b, ..] = &array;
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LL + let [a, b, ..] = &array;
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error: dereferencing a slice pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:41:12
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LL | if let &[ref a, ref b] = slice {}
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| ^^^^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &[ref a, ref b] = slice {}
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LL + if let [a, b] = slice {}
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error: dereferencing a slice pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:42:12
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LL | if let &[ref a, ref b] = &vec[..] {}
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| ^^^^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &[ref a, ref b] = &vec[..] {}
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LL + if let [a, b] = &vec[..] {}
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error: dereferencing a slice pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:44:12
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LL | if let &[ref a, ref b, ..] = slice {}
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| ^^^^^^^^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &[ref a, ref b, ..] = slice {}
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LL + if let [a, b, ..] = slice {}
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error: dereferencing a slice pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:45:12
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LL | if let &[ref a, .., ref b] = slice {}
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| ^^^^^^^^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &[ref a, .., ref b] = slice {}
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LL + if let [a, .., b] = slice {}
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error: dereferencing a slice pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:46:12
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LL | if let &[.., ref a, ref b] = slice {}
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| ^^^^^^^^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &[.., ref a, ref b] = slice {}
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LL + if let [.., a, b] = slice {}
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error: dereferencing a slice pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:48:12
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LL | if let &[ref a, _] = slice {}
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| ^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &[ref a, _] = slice {}
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LL + if let [a, _] = slice {}
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error: dereferencing a tuple pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:50:12
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LL | if let &(ref a, ref b, ref c) = &tuple {}
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| ^^^^^^^^^^^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &(ref a, ref b, ref c) = &tuple {}
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LL + if let (a, b, c) = &tuple {}
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error: dereferencing a tuple pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:51:12
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LL | if let &(ref a, _, ref c) = &tuple {}
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| ^^^^^^^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &(ref a, _, ref c) = &tuple {}
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LL + if let (a, _, c) = &tuple {}
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error: dereferencing a tuple pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:52:12
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LL | if let &(ref a, ..) = &tuple {}
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| ^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &(ref a, ..) = &tuple {}
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LL + if let (a, ..) = &tuple {}
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error: dereferencing a tuple pattern where every element takes a reference
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--> tests/ui/needless_borrowed_ref.rs:54:12
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LL | if let &TupleStruct(ref a, ..) = &tuple_struct {}
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| ^^^^^^^^^^^^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &TupleStruct(ref a, ..) = &tuple_struct {}
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LL + if let TupleStruct(a, ..) = &tuple_struct {}
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error: dereferencing a struct pattern where every field's pattern takes a reference
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--> tests/ui/needless_borrowed_ref.rs:56:12
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LL | if let &Struct {
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| ____________^
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LL | | ref a,
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LL | | b: ref b,
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LL | | c: ref renamed,
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LL | | } = &s
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| |_____^
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help: try removing the `&` and `ref` parts
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LL ~ if let Struct {
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LL ~ a,
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LL ~ b: b,
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LL ~ c: renamed,
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error: dereferencing a struct pattern where every field's pattern takes a reference
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--> tests/ui/needless_borrowed_ref.rs:63:12
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LL | if let &Struct { ref a, b: _, .. } = &s {}
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| ^^^^^^^^^^^^^^^^^^^^^^^^^^^
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help: try removing the `&` and `ref` parts
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LL - if let &Struct { ref a, b: _, .. } = &s {}
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LL + if let Struct { a, b: _, .. } = &s {}
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error: aborting due to 17 previous errors
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