Every time I think I understand generics better (and can answer without compiling), I get to an example where this theory breaks. Here is a very simple example: And two invocations: To me, none of the invocations should compile. A String is not a supertype of List. Still, the second one compiles. But let's suppose that this happens because the

Java generics impossible assignment?

Every time I think I understand generics better (and can answer without compiling), I get to an example where this theory breaks. Here is a very simple example:

static void consumer(List<? super List<String>> param) {
    System.out.println(param);
}

JavaScript
​x
 
static void consumer(List<? super List<String>> param) {    System.out.println(param);}​

And two invocations:

public static void main(String[] args) {
    List<String> list = List.of("123");
    consumer(list);
    consumer(List.of("123"));
}

JavaScript
 
public static void main(String[] args) {    List<String> list = List.of("123");    consumer(list);    consumer(List.of("123"));}​

To me, none of the invocations should compile. A String is not a supertype of List. Still, the second one compiles. But let’s suppose that this happens because the compiler could infer some type here. Of course such a type does not exist and it will fail at runtime, right? Right? Nope. It just works. As such, can someone bring some sanity to my life please?

Answer

Ah darn!

javac  --debug=verboseResolution=all Sandbox.java

JavaScript
 
javac  --debug=verboseResolution=all Sandbox.java​

shows that consumer(List.of("123")) is compiled to:

instantiated signature: (Object)List<Object>
target-type: List<? super List<String>>

JavaScript
 
instantiated signature: (Object)List<Object>target-type: List<? super List<String>>​

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Answer