Refine special bridge generating rule

Generate special bridge even in case current class has Kotlin superclass implementing
this builtin method, but that super class was generated without special bridge
(e.g. because it would have the same signature)

 #KT-9901 Fixed
This commit is contained in:
Denis Zharkov
2015-11-12 14:08:18 +03:00
parent abf7ae547e
commit f5a086140e
4 changed files with 86 additions and 16 deletions
@@ -0,0 +1,59 @@
var result = ""
public abstract class AbstractFoo<K, V> : Map<K, V> {
override operator fun get(key: K): V? {
result = "AbstractFoo"
return null
}
override val size: Int
get() = throw UnsupportedOperationException()
override fun isEmpty(): Boolean {
throw UnsupportedOperationException()
}
override fun containsKey(key: K): Boolean {
throw UnsupportedOperationException()
}
override fun containsValue(value: V): Boolean {
throw UnsupportedOperationException()
}
override val keys: Set<K>
get() = throw UnsupportedOperationException()
override val values: Collection<V>
get() = throw UnsupportedOperationException()
override val entries: Set<Map.Entry<K, V>>
get() = throw UnsupportedOperationException()
}
public open class StringFoo<E> : AbstractFoo<String, E>() {
override operator fun get(key: String): E? {
result = "StringFoo"
return null
}
}
public class IntFoo<E> : AbstractFoo<Int, E>() {
override operator fun get(key: Int): E? {
result = "IntFoo"
return null
}
}
public class AnyFoo<E> : AbstractFoo<Any?, E>() {}
fun box(): String {
StringFoo<String>().get("")
if (result != "StringFoo") return "fail 1: $result"
IntFoo<String>().get(1)
if (result != "IntFoo") return "fail 2: $result"
AnyFoo<String>().get(null)
if (result != "AbstractFoo") return "fail 3: $result"
return "OK"
}