Differencebetween the | and || Operators

What is the difference between the | and || or operators?

Just like the & and && operator, the double Operator is a "short-circuit" operator.

For example:

if(condition1 || condition2 || condition3)

If condition1 is true, condition 2 and 3 will NOT be checked.

if(condition1 | condition2 | condition3)

This will check conditions 2 and 3, even if 1 is already true. As your conditions can be quite expensive functions, you can get a good performance boost by using them.

There is one big caveat, NullReferences or similar problems. For example:

if(class != null && class.someVar < 20)

If class is null, the if-statement will stop after class != null is false. If you only use &, it will try to check class.someVar and you get a nice NullReferenceException. With the Or-Operator that may not be that much of a trap as it's unlikely that you trigger something bad, but it's something to keep in mind.

No one ever uses the single & or | operators though, unless you have a design where each condition is a function that HAS to be executed. Sounds like a design smell, but sometimes (rarely) it's a clean way to do stuff. The & operator does "run these 3 functions, and if one of them returns false, execute the else block", while the | does "only run the else block if none return false" - can be useful, but as said, often it's a design smell.

There is a Second use of the | and & operator though: Bitwise Operations.

Difference between || and ?? operators

The main difference is that nullish coalescing(??) operator will only give the result as the right operand only if the left operand is either null or undefined.

Whereas the OR(||) operator will give the result as right operand for all the falsy values of the left operand.

Below are some examples

• Snippet 1: With 0 as input

const a = 0;
// a || 10 --> Will result in 10, as || operator considers 0 as falsy value and resulting the right side operand
console.log(`a || 10 = ${a || 10}`);
// a ?? 10 --> Will result in 0, as ?? operator considers 0 as truthy value and resulting the left side operand
console.log(`a ?? 10 = ${a ?? 10}`);

What is the difference between the | and || operators?

| is a bitwise or, || is a boolean or.

What's the difference between ( | ) and ( || )?

| is a bitwise or, || is a logical or.

A bitwise or takes the two numbers and compares them on a bit-by-bit basis, producing a new integer which combines the 1 bits from both inputs. So 0101 | 1010 would produce 1111.

A logical or || checks for the "truthiness" of a value (depends on the type, for integers 0 is false and non-zero is true). It evaluates the statement left to right, and returns the first value which is truthy. So 0101 || 1010 would return 0101 which is truthy, therefore the whole statement is said to be true.

The same type of logic applies for & vs &&. 0101 & 1010 = 0000. However 0101 && 1010 evaluates to 1010 (&& returns the last truthy value so long as both operands are truthy).

Why do we usually use || over |? What is the difference?

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.

It's a matter of if you want to short-circuit the evaluation or not -- most of the time you want to.

A good way to illustrate the benefits of short-circuiting would be to consider the following example.

Boolean b = true;
if(b || foo.timeConsumingCall())
{
   //we entered without calling timeConsumingCall()
}

Another benefit, as Jeremy and Peter mentioned, for short-circuiting is the null reference check:

if(string != null && string.isEmpty())
{
    //we check for string being null before calling isEmpty()
}

more info

Php: What is the difference between | and || operator

| is a bitwise or, || is a logical or. | operates on binary values whereas || operates on boolean ones.

E.g. 5 | 3 is 0101 OR 0011 which is 0111 which is 7, whereas True || False is True and False || False is False.

what is the difference between `| `and` || `in java?

a || b never evaluates b if a evaluates to true.

What do the & and | operators do? How are they different from && and ||? Swift

See the Wikipedia page on Bitwise Operation and the Swift documentation for bitwise operators.

These are bitwise operators. & is bitwise AND and | is bitwise OR.

See these examples:

    0011 (decimal 3)
AND 0010 (decimal 2)
  = 0010 (decimal 2)

   0101 (decimal 5)
OR 0011 (decimal 3)
 = 0111 (decimal 7)

Source: Wikipedia

The uses of bitwise operators have been discussed before on StackOverflow:

• practical applications of bitwise operations
• Using bitwise operations

A use of bitwise XOR (not in your question, but a cool logic gate anyway) that caught my eye (by @Vilx- here) (I don't know how it works, but the answer was accepted and up-voted 34 times)

EDIT: If you want to know how it works, there's a nice simple proof over on XOR swap algorithm - Wikipedia

Swapping two integer variables without an intermediary variable:

A = A^B // A is now XOR of A and B
B = A^B // B is now the original A
A = A^B // A is now the original B

If these don't help, the Wikipedia page I already linked to twice in this post has an Applications section, but they don't really apply to higher-level languages (unless for some reason you want to optimize your arithmetic to use only bitwise operations).

---

Related Topics