Why is my if statement behaving this way?

Question

I ran across this puzzle today. Obviously, this isn't correct style, but I'm still curious as to why no output is coming out.

int x = 9;
int y = 8;
int z = 7;

if (x > 9) if (y > 8) System.out.println("x > 9 and y > 8");

else if (z >= 7) System.out.println("SHOULD OUTPUT THIS x <= 9 and z >= 7");

else
  System.out.println("x <= 9 and z < 7");

The above has no output when run. But, when we add in brackets for the if-statement, suddenly the logic behaves as I expect.

int x = 9;
int y = 8;
int z = 7;

if (x > 9) {
  if (y > 8) System.out.println("x > 9 and y > 8");
}

else if (z >= 7) System.out.println("SHOULD OUTPUT THIS x <= 9 and z >= 7");

else
  System.out.println("x <= 9 and z < 7");

This outputs "SHOULD OUTPUT THIS x <= 9 and z >= 7". What is going on here?

Thanks!

Answer 1

If you rewrite the first way like this (which is how it is behaving), it is easier to understand

if (x > 9)
  if (y > 8) System.out.println("x > 9 and y > 8");
  else if (z >= 7) System.out.println("SHOULD OUTPUT THIS x <= 9 and z >= 7");
  else
    System.out.println("x <= 9 and z < 7");

Since x is not > 9, the block never executes.

Answer 2

This:

if (x > 9) ... if (y > 8) ... else if (z >= 7) ... else

is ambiguous, because during parsing the else could be bound to the first if or the second if. (This is called the dangling else problem). The way Java (and many other languages) deals with this is to make the first meaning illegal, so the else clauses always bind to the innermost if statements.