Description Viliam Anirud 2016-08-17 16:51:46 EDT

When converting lambda expression to method reference or vice versa and when the method is an instance method, the instance scope is not taken into account.

   myInstance::myMethod

is converted to

   () -> myInstance.myMethod()

However, in the former case, myInstance is evaluated at the lambda creation time while in the later case it is evaluated when the lambda is executed. These two are not semantically equivalent. The above conversion is only safe, if myInstance is final. The lambda expression should be:

   MyInstanceClass myInstanceCopy = myInstance;
   () -> myInstanceCopy.myMethod()

Here is a more complete example:


  import java.util.function.Supplier;
  
  public class Test {
  
    static class Clazz {
      private final String value;
      
      public Clazz(String value) {
        this.value = value;
      }
      
      public String getValue() {
        return value;
      }
    }
    
    static Clazz myInstance;
    
    public static void main(String[] args) {
      myInstance = new Clazz("a");
      
      Supplier<String> supplierMethodRef = myInstance::getValue;
      // The following lambda expression is the result of conversion
      // of the method reference on the previous line to lambda
      Supplier<String> supplierLambda = () -> myInstance.getValue();
  
      System.out.println(supplierLambda.get());
      System.out.println(supplierMethodRef.get());
      
      // after changing the value of "x", the result is different in each case
      myInstance = new Clazz("b");
      System.out.println(supplierLambda.get());
      System.out.println(supplierMethodRef.get());
    }
  }



As far as I know, there is not a way to express a lambda as a method reference in equivalent way, if the instance is not final. The current implementation is confusing, as people believe that refactored code is equivalent.