I have this question in my beginner Java programming class and I can't find in my book what a single '|' means as an operator.
The question is:
int j = 0;
if ((8 > 4) | (j++ == 7))
System.out.println("j = " + j);
Is j = 1?
Explain why.
The book explains the OR operator "||" with examples but it doesn't show this single "|". Does the meaning of the operator change between the two?
Yes, the meaning changes, || is the logical or operator, while | is the bitwise or.
Since this operator operates (...) at the bit level, it doesn't short circuit the same way as ||, meaning, even if the first operand is true, the second operand will be evaluated.
If you use the || and && forms, rather than the | and & forms of these operators, Java will not bother to evaluate the right-hand operand alone.
Boolean b = true;
if(b || foo.test())
{
//we entered without calling test()
}
if(b | foo.test())
{
//we entered calling test()
}
| -> This is bit-wise OR operator performs a bitwise inclusive OR operation and perform the both operand(right and left).
The bitwise OR "|" operator produces 1 if either one or both of the bits in its operands are 1. However, if both of the bits are 0 then this operator produces 0. To be more precise OR "|" operator returns 1 in all cases except when both the bits of both the operands are 0.
if ((8 > 4) | (j++ == 7)) System.out.println("j = " + j);
Here 8>4 will return true and j++ == 7 will return false. so it will be if(true|false) and here bit-wise operator will return true or 1 and print value of j which became 1 after j++.
OR conditional operator.The | operator is bitwise OR, but you may ask why would someone use the bitwise OR on boolean variables?
Logical OR operator (||) has a feature called short-circuiting, which means that if the left operand evaluates to true, no further processing is done, and the right hand operand even will not evaluate. But bitwise OR operator | always evaluates both operands, and if applied on boolean operands, evaluates both of them and then applies a simple or on the results.
In your case, (j++ == 7) is always evaluated, regardless of the result of 8 > 4. This does not happen with || operator. So j will be incremented by one and the output of the program is 1.
The answer is No because j is initialized at 0 but is never assigned after this. So the value of j stays at 0 for the whole time.
