What kind of lambda expression syntax is this in Java 8?

Question

This is from one question about lambda expression. I'm baffled with the syntax in line check((h, l) -> h > l, 0); :

The check() function requires a Climb object and an int. The line above does not provide any Climb object. And what does h > l, 0 mean?

interface Climb {
  boolean isTooHigh(int height, int limit);
}

class Climber {
  public static void main(String[] args) {
    check((h, l) -> h > l, 0);
  }
  private static void check(Climb climb, int height) {
    if (climb.isTooHigh(height, 1)) 
      System.out.println("too high");
    else 
      System.out.println("ok");
  }
}

Answer 1

Your Climb interface respects the contract of a functional interface , that is, an interface with a single abstract method.

Hence, an instance of Climb can be implemented using a lambda expression, that is, an expression that takes two ints as parameters and returns a boolean in this case.

(h, l) -> h > l is the lambda expression that implements it. h and l are the parameters ( int ) and it will return whether h > l (so the result is indeed a boolean ). So for instance you could write:

Climb climb = (h, l) -> h > l;
System.out.println(climb.isTooHigh(0, 2)); //false because 0 > 2 is not true

Answer 2

Apparently, (h, l) -> h > l is a lambda expression where the result is of a boolean type and the 0 is the second argument of check , which is unrelated to the lambda expression itself; the 0 does not belong to the lambda expression as such.

Answer 3

(h, l) -> h > l is the climb object. It is a lambda that returns true when the first argument (h) is greater than the second (l). 0 is the int, and the comma separates the arguments.