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28cc919fb4
When a specialization of a generic type occurs within the signature of
a generic function, it implies that the generic arguments meet the
requirements of the generic type. For example, in
func printHashes<K, V>(dict : Dictionary<K, V>) {
for (k, v) in dict {
print("\(k.hashValue())\n")
}
}
the presence of Dictionary<K, V> in the signature implies that K and V
meet the requirements on Dictionary's generic parameters, i.e., that K
is Hashable. Thus, infer that K is Hashable in printHashes().
Fixes the easy part of <rdar://problem/14691708>. Same-type and
superclass requirements are more interesting.
<rdar://problem/14691708>
Swift SVN r7574
48 lines
946 B
Swift
48 lines
946 B
Swift
// RUN: %swift -parse %s -verify
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protocol P1 {
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func p1()
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}
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struct X1<T : P1> {
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func getT() -> T { }
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}
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class X2<T : P1> {
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func getT() -> T { }
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}
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class X3 { }
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struct X4<T : X3> {
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func getT() -> T { }
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}
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// Infer protocol requirements from the parameter type of a generic function.
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func inferFromParameterType<T>(x : X1<T>) {
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x.getT().p1()
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}
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// Infer protocol requirements from the return type of a generic function.
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func inferFromReturnType<T>(x : T) -> X1<T> {
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x.p1()
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}
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// Infer protocol requirements from the superclass of a generic parameter.
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func inferFromSuperclass<T, U : X2<T>>(t : T, u : U) -> T {
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t.p1()
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}
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// Infer protocol requirements from the parameter type of a constructor.
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struct InferFromConstructor {
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constructor<T> (x : X1<T>) {
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x.getT().p1()
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}
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}
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// FIXME: Infer superclass requirements.
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// FIXME: Infer same-type requirements, which requires us to process
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// the where clause in its entirety.
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