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49277e4e05
foo(_: 3) is equivalent to foo(3), so calling a function that has _ as an argument label (func foo(`_`: 3)) still requires the _ to be escaped. Before this patch, the compiler would suggest removing the `s, even though that changes behaviour. Fixes rdar://problem/31077797.
58 lines
1.2 KiB
Swift
58 lines
1.2 KiB
Swift
// RUN: %target-typecheck-verify-swift
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struct SomeRange { }
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// Function declarations.
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func paramName(_ func: Int, in: SomeRange) { }
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func firstArgumentLabelWithParamName(in range: SomeRange) { }
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func firstArgumentLabelWithParamName2(range in: SomeRange) { }
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func escapedInout(`inout` value: SomeRange) { }
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struct SomeType {
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// Initializers
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init(func: () -> ()) { }
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init(init func: () -> ()) { }
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// Subscripts
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subscript (class index: AnyClass) -> Int {
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return 0
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}
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subscript (class: AnyClass) -> Int {
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return 0
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}
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subscript (struct: Any.Type) -> Int {
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return 0
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}
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}
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class SomeClass { }
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// Function types.
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typealias functionType = (_ in: SomeRange) -> Bool
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// Calls
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func testCalls(_ range: SomeRange) {
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paramName(0, in: range)
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firstArgumentLabelWithParamName(in: range)
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firstArgumentLabelWithParamName2(range: range)
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var st = SomeType(func: {})
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st = SomeType(init: {})
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_ = st[class: SomeClass.self]
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_ = st[SomeClass.self]
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_ = st[SomeType.self]
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escapedInout(`inout`: range)
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// Fix-Its
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paramName(0, `in`: range) // expected-warning{{keyword 'in' does not need to be escaped in argument list}}{{16-17=}}{{19-20=}}
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}
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// rdar://problem/31077797
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func foo(`_`: Int) {}
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foo(`_`: 3)
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let f = foo(`_`:)
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f(3)
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