Wildcards in generics behave differently in Java 7 and 8

I have the code compiled when it is compiled into Java 7 (eclipse compiler), but it fails when I set the project to Java 8:

package scratch;

class Param<T extends Comparable<T>> {
  public Comparable<?> get() {
    return null;
  }
}

public class Condition<T extends Comparable<T>> {
  public static <T extends Comparable<T>> Condition<T> isInRange(T lower,T upper) {
    return null;
  }

  public void foo() {
    Comparable bound = null;                 // Line 15 
    Param<?> param = new Param<Double>();
    Condition.isInRange(param.get(),bound); // Line 17
  }
}

In Java 7, I received the following warning:

>Line 15: comparable is the original type We should parameterize the reference to generic type Comparable > seventeenth lines: type safety: isInRange (Comparable, Comparable) of isInRange (T, T), which is not a general method of calling type.

When I add On line 15, the warning disappears, but I get an error on line 17:

Who knows what led to this intransigence?

PS: I added these ugly casts to compile code under two versions of Java:

Condition.isInRange((Comparable)param.get(),(Comparable) bound);

Solution

First of all, I'm not a generic expert, so my answer may not be correct I think the problem is how the isinrange method is defined

<T extends Comparable<T>>

This is a recursive definition and is not a problem in itself Look at courses like double and long I believe that because of how to resolve recursive definitions, the type of method (isinrange) must be a class that implements comparable, and it uses itself as a generic type of comparable; For example, class C implements comparable < C >