Difference Between the | and || or Operators

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}`);

Difference between OR operator || and | in Java?

This is simple. http://www.roseindia.net/help/java/o/java-operator.shtml says:

OR operator is a kind of a conditional operators, which is represented
by | symbol. It returns either true or false value based on the state
of the variables i.e. the operations using conditional operators are
performed between the two boolean expressions.

The OR operator (|) is similar to the Conditional-OR operator (||) and
returns true, if one or another of its operand is true.

Note: In || operator if have more than one condition and if first condition return true then other conditions ignored but in | operator all condition examin.

There are more information on http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html

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).

What is difference between logical OR and short circuit OR operator?

The difference is that the short circuit operator doesn't evaluate the second operand if the first operand is true, which the logical OR without short circuit always evaluates both operands.

You wouldn't see any difference in your simple test, since both should give the same output assuming no exception is thrown, but if you try something like this :

String s = null;
System.out.println("a || b = " + s==null || s.length() == 0 );
System.out.println("a | b = " + s==null | s.length() == 0 );

The first operator will give you true, while the second operator will give you NullPointerException, since only the | operator will attempt to evaluate s.length() == 0.

Difference between | and || , or & and &&

The operators |, &, and ~ act on individual bits in parallel. They can be used only on integer types. a | b does an independent OR operation of each bit of a with the corresponding bit of b to generate that bit of the result.

The operators ||, &&, and ! act on each entire operand as a single true/false value. Any data type can be used that implicitly converts to bool. Many data types, including float implicitly convert to bool with an implied !=0 operation.

|| and && also "short circuit". That means whenever the value of the result can be determined by just the first operand, the second is not evaluated. Example:

ptr && (*ptr==7) If ptr is zero, the result is false without any risk of seg faulting by dereferencing zero.

You could contrast that with (int)ptr & (*ptr). Ignoring the fact that this would be a bizarre operation to even want, if (int)ptr were zero, the entire result would be zero, so a human might think you don't need the second operand in that case. But the program will likely compute both anyway.

Difference between | and || or & and && for comparison

in C (and other languages probably) a single | or & is a bitwise comparison.

The double || or && is a logical comparison.

Edit: Be sure to read Mehrdad's comment below regarding "without short-circuiting"

In practice, since true is often equivalent to 1 and false is often equivalent to 0, the bitwise comparisons can sometimes be valid and return exactly the same result.

There was once a mission critical software component I ran a static code analyzer on and it pointed out that a bitwise comparison was being used where a logical comparison should have been. Since it was written in C and due to the arrangement of logical comparisons, the software worked just fine with either. Example:

if ( (altitide > 10000) & (knots > 100) )
...

What's the difference between or and | in ruby?

The | operator is a binary mathematical operator, that is it does a binary OR and works on a numerical level:

1 | 2
# => 3
4 | 3
# => 7
1 | 2 | 3
# => 3

This is because it's manipulating individual values as if they were binary:

0b01 | 0b10
# => 3 (0b11)

The || operator is a logical one, that is it returns the first value that's logically true. In Ruby only literal nil and false values evaluates as logically false, everything else, including 0, empty strings and arrays, is true.

So:

1 || 2
# => 1
0 || 1
# => 0

The or operator works almost exactly the same as || except it's at a much lower precedence. That means other operators are evaluated first which can lead to some problems if you're not anticipating this:

a = false || true
# => true
a
# => true

a = false or true
# => true
a
# => false

This is because it's actually interpreted as:

(a = false) or true

This is because = has a higher precedence when being evaluated.

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