Are short-circuit operators overloaded?

4
votes

There are two cases of using bitwise operator:

For boolean

boolean a = true;
boolean b= false;
boolean c = a|b; // Giving response after logical OR for booleans.

For integer

int a = 10;
int b = 20;
int c = a|b;  // Giving response after bitwise OR for boolean equivalents of "a" and "b".

Both of the above cases are in compliance of http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.22.2.

Is the operator | overloaded?

I just intent to ask a very simple question: Is "|" overloaded or it performs same task of bitwise OR for both booleans (binary equivalents of course) and integer?

2
javaor-operator
What would the difference between a bitwise and a logical OR be for a boolean? - millimoose
Also I'm not sure of you're asking. Is | overloaded? Clearly it is insofar as it can be applied to different operand types, but you're saying as much in your question. Are you asking what the difference between the two is? Well you have the JLS open so it's right there, but | isn't short-circuiting. Are you asking about how to do bitwise-OR on booleans? That distinction doesn't make sense for a data type that represents one bit. - millimoose
Why not bitwise operation on booleans makes sense? Booleans surely have some binary equivalents and bitwise OR for them may provide us the desired results. - Amber
The binary equivalent of a boolean is one bit. This is a lie because you can only straightforwardly work with whole registers at the CPU level, but Java more or less maintains the illusion - you can't directly cast between booleans and integers or otherwise use booleans as arithmetic values. - millimoose
There are no short-circuit operators here. The short-circuit operators are || and &&. - user207421

2 Answers

4
votes

You might find an answer here:

When both operands of a &, ^, or | operator are of type boolean or Boolean, then the type of the bitwise operator expression is boolean. In all cases, the operands are subject to unboxing conversion (ยง5.1.8) as necessary.

If you use the | operator on boolean, then the result would be like using || but note that the difference is in the case of:

boolean a = true, b = false;
boolean c = a | b; //b will still be evaluated
c = a || b;        //b will not be evaluated

I'm not sure what do mean by asking if it is overloaded, because since it can be used on different types, it is overloaded.

2
votes

I am not completely sure what you are asking, but at least the bytecode is different between boolean c = a|b and boolean c = a||b:

boolean a = true;
boolean b = false;

boolean c = a|b;

ILOAD 1
ILOAD 2
IOR
ISTORE 3

boolean c = a||b;

ILOAD 1
IFNE L4
ILOAD 2
IFNE L4
ICONST_0

So the two operators effectively result in different operations on a bytecode level. Most specifically, || only evaluates the second operand if the first operand is false, while | evaluates both operands in any case:

public boolean a() {
  System.out.println("   a");
  return true;
}

public boolean b() {
  System.out.println("   b");
  return false;
}

public void c() {
  System.out.println("a() | b()");
  boolean r1 = a() | b();
  System.out.println("\na() || b()");
  boolean r2 = a() || b();
}

Output:

a() | b()
   a
   b

a() || b()
   a

At the same time, the bytecode for integer bitwise or is the same as for boolean bitwise or:

int a1 = 10;
int a2 = 20;
int c1 = a1 | a2;

ILOAD 4
ILOAD 5
IOR
ISTORE 6