How Java handles IF statements and efficiency

Question

I'm just curious about how Java actually works when it come to if statements. (Note: when I say "component" below I mean the idividual parts checked by the statement, e.g. a, b, c)

Which is more efficient in terms of calculations?

if (a && b && c) { do stuff }

or

if (a) {
  if (b) {
    if (c) {
    do stuff }
  }
}

The reason why I ask is because it's important what Java does in the first version. Does it check every single thing in the statement or does it check a and if it is false then cancel checking the rest of the statement?

If this is the case then it makes sense to put the component most likely to fail as the first component in the statement.

If the whole statement is checked every time then it makes more sense to split the components into a bunch of different statements, as in the second example.

Answer 1

In Java, && and || are guaranteed to short-circuit: the operands are evaluated left-to-right, and the evaluation stops as soon as the result is known with certainty.

From the JLS:

The && operator is like & (§15.22.2), but evaluates its right-hand operand only if the value of its left-hand operand is true. It is syntactically left-associative (it groups left-to-right).

The || operator is like | (§15.22.2), but evaluates its right-hand operand only if the value of its left-hand operand is false. It is syntactically left-associative (it groups left-to-right).

This means that the two code snippets in your question are exactly equivalent.

Answer 2

These code snippets are completely equivalent and produce exactly the same bytecode:

0:  iload_1
1:  ifeq    20 //a == false (0)
4:  iload_2
5:  ifeq    20 //b == false (0)
8:  iload_3
9:  ifeq    20 //c == false (0)
12: //do stuff
20: return

Although the explanation is different, the code generator produces the same output. In the first case due to lazy evaluation subsequent terms aren't evaluated if previous one is false. In the second case the runtime won't go deeper if surrounding if condition is not met.