Dmitriy Novozhilov
d4f55894b4
If some function is not fake-override, then its type should be just default type of containing class For fake overrides the default type calculated in the following way: 1. Find first overridden function, which is not fake override 2. Take its containing class 3. Find supertype of current containing class with type constructor of class from step 2 ^KT-60252 Fixed
52 lines
991 B
Kotlin
Vendored
52 lines
991 B
Kotlin
Vendored
// FIR_IDENTICAL
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import C.f
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import C.p
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import C.ext
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import C.g1
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import C.g2
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import C.fromClass
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import C.fromInterface
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import C.genericFromSuper
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interface I<G> {
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fun <T> T.fromInterface(): T = this
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fun genericFromSuper(g: G) = g
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}
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open class BaseClass {
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val <T> T.fromClass: T
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get() = this
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}
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object C: BaseClass(), I<String> {
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fun f(s: Int) = 1
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fun f(s: String) = 2
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fun Boolean.f() = 3
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var p: Int = 4
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val Int.ext: Int
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get() = 6
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fun <T> g1(t: T): T = t
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val <T> T.g2: T
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get() = this
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}
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fun box(): String {
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if (f(1) != 1) return "1"
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if (f("s") != 2) return "2"
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if (true.f() != 3) return "3"
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if (p != 4) return "4"
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p = 5
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if (p != 5) return "5"
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if (5.ext != 6) return "6"
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if (g1("7") != "7") return "7"
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if ("8".g2 != "8") return "8"
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if (9.fromInterface() != 9) return "9"
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if ("10".fromClass != "10") return "10"
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if (genericFromSuper("11") != "11") return "11"
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return "OK"
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}
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