What is the difference between & and && in Java?

Question

I always thought that && operator in Java is used for verifying whether both its boolean operands are true, and the & operator is used to do Bit-wise operations on two integer types.

Recently I came to know that & operator can also be used verify whether both its boolean operands are true, the only difference being that it checks the RHS operand even if the LHS operand is false.

Is the & operator in Java internally overloaded? Or is there some other concept behind this?

Answer 1

& <-- verifies both operands
&& <-- stops evaluating if the first operand evaluates to false since the result will be false

(x != 0) & (1/x > 1) <-- this means evaluate (x != 0) then evaluate (1/x > 1) then do the &. the problem is that for x=0 this will throw an exception.

(x != 0) && (1/x > 1) <-- this means evaluate (x != 0) and only if this is true then evaluate (1/x > 1) so if you have x=0 then this is perfectly safe and won't throw any exception if (x != 0) evaluates to false the whole thing directly evaluates to false without evaluating the (1/x > 1).

EDIT:

exprA | exprB <-- this means evaluate exprA then evaluate exprB then do the |.

exprA || exprB <-- this means evaluate exprA and only if this is false then evaluate exprB and do the ||.

Answer 2

Besides not being a lazy evaluator by evaluating both operands, I think the main characteristics of bitwise operators compare each bytes of operands like in the following example:

int a = 4;
int b = 7;
System.out.println(a & b); // prints 4
//meaning in an 32 bit system
// 00000000 00000000 00000000 00000100
// 00000000 00000000 00000000 00000111
// ===================================
// 00000000 00000000 00000000 00000100

Answer 3

boolean a, b;

Operation     Meaning                       Note
---------     -------                       ----
   a && b     logical AND                    short-circuiting
   a || b     logical OR                     short-circuiting
   a &  b     boolean logical AND            not short-circuiting
   a |  b     boolean logical OR             not short-circuiting
   a ^  b     boolean logical exclusive OR
  !a          logical NOT

short-circuiting        (x != 0) && (1/x > 1)   SAFE
not short-circuiting    (x != 0) &  (1/x > 1)   NOT SAFE

Answer 4

It depends on the type of the arguments...

For integer arguments, the single ampersand ("&")is the "bit-wise AND" operator. The double ampersand ("&&") is not defined for anything but two boolean arguments.

For boolean arguments, the single ampersand constitutes the (unconditional) "logical AND" operator while the double ampersand ("&&") is the "conditional logical AND" operator. That is to say that the single ampersand always evaluates both arguments whereas the double ampersand will only evaluate the second argument if the first argument is true.

For all other argument types and combinations, a compile-time error should occur.

Answer 5

&& is a short circuit operator whereas & is a AND operator.

Try this.

    String s = null;
    boolean b = false & s.isEmpty(); // NullPointerException
    boolean sb = false && s.isEmpty(); // sb is false

Answer 6

I think my answer can be more understandable:

There are two differences between & and &&.

If they use as logical AND

& and && can be logical AND, when the & or && left and right expression result all is true, the whole operation result can be true.

when & and && as logical AND, there is a difference:

when use && as logical AND, if the left expression result is false, the right expression will not execute.

Take the example :

String str = null;

if(str!=null && !str.equals("")){  // the right expression will not execute
        
}

If using &:

String str = null;

if(str!=null & !str.equals("")){  // the right expression will execute, and throw the NullPointerException 
        
}

An other more example:

int x = 0;
int y = 2;
if(x==0 & ++y>2){
    System.out.print(“y=”+y);  // print is: y=3
}

---

int x = 0;
int y = 2;
if(x==0 && ++y>2){
    System.out.print(“y=”+y);  // print is: y=2
}

& can be used as bit operator

& can be used as Bitwise AND operator, && can not.

The bitwise AND " &" operator produces 1 if and only if both of the bits in its operands are 1. However, if both of the bits are 0 or both of the bits are different then this operator produces 0. To be more precise bitwise AND " &" operator returns 1 if both of the two bits is 1 and it returns 0 if any of the bits is 0. 

From the wiki page:

http://www.roseindia.net/java/master-java/java-bitwise-and.shtml

Answer 7

it's as specified in the JLS (15.22.2):

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 (§5.1.8) as necessary.

For &, the result value is true if both operand values are true; otherwise, the result is false.

For ^, the result value is true if the operand values are different; otherwise, the result is false.

For |, the result value is false if both operand values are false; otherwise, the result is true.

The "trick" is that & is an Integer Bitwise Operator as well as an Boolean Logical Operator. So why not, seeing this as an example for operator overloading is reasonable.

Answer 8

‘&&’ : - is a Logical AND operator produce a boolean value of true or false based on the logical relationship of its arguments.

For example: - Condition1 && Condition2

If Condition1 is false, then (Condition1 && Condition2) will always be false, that is the reason why this logical operator is also known as Short Circuit Operator because it does not evaluate another condition. If Condition1 is false , then there is no need to evaluate Condtiton2.

If Condition1 is true, then Condition2 is evaluated, if it is true then overall result will be true else it will be false.

‘&’ : - is a Bitwise AND Operator. It produces a one (1) in the output if both the input bits are one. Otherwise it produces zero (0).

For example:-

int a=12; // binary representation of 12 is 1100

int b=6; // binary representation of 6 is 0110

int c=(a & b); // binary representation of (12 & 6) is 0100

The value of c is 4.

for reference , refer this http://techno-terminal.blogspot.in/2015/11/difference-between-operator-and-operator.html

Answer 9

With booleans, there is no output difference between the two. You can swap && and & or || and | and it will never change the result of your expression.

The difference lies behind the scene where the information is being processed. When you right an expression "(a != 0) & ( b != 0)" for a= 0 and b = 1, The following happens:

left side: a != 0 --> false
right side: b 1= 0 --> true
left side and right side are both true? --> false
expression returns false

When you write an expression (a != 0) && ( b != 0) when a= 0 and b = 1, the following happens:

a != 0 -->false
expression returns false

Less steps, less processing, better coding, especially when doing many boolean expression or complicated arguments.

Answer 10

&& and || are called short circuit operators. When they are used, for || - if the first operand evaluates to true, then the rest of the operands are not evaluated. For && - if the first operand evaluates to false, the rest of them don't get evaluated at all.

so if (a || (++x > 0)) in this example the variable x won't get incremented if a was true.

Answer 11

Besides && and || being short circuiting, also consider operator precedence when mixing the two forms. I think it will not be immediately apparent to everybody that result1 and result2 contain different values.

boolean a = true;
boolean b = false;
boolean c = false;

boolean result1 = a || b && c; //is true;  evaluated as a || (b && c)
boolean result2 = a  | b && c; //is false; evaluated as (a | b) && c

Answer 12

& is a bitwise operator plus used for checking both conditions because sometimes we need to evaluate both condition. But && logical operator go to 2nd condition when first condition give true.

Answer 13

all answers are great, and it seems that no more answer is needed but I just wonted to point out something about && operator called dependent condition

In expressions using operator &&, a condition—we’ll call this the dependent condition—may require another condition to be true for the evaluation of the dependent condition to be meaningful.

In this case, the dependent condition should be placed after the && operator to prevent errors.

Consider the expression (i != 0) && (10 / i == 2). The dependent condition (10 / i == 2) must appear after the && operator to prevent the possibility of division by zero.

another example (myObject != null) && (myObject.getValue() == somevaluse)

and another thing: && and || are called short-circuit evaluation because the second argument is executed or evaluated only if the first argument does not suffice to determine the value of the expression

References: Java™ How To Program (Early Objects), Tenth Edition

Answer 14

In respect of the AND and OR operators, Java has got two types of evaluation namely Short-Circuit evaluation and full evaluation.

&& || Short-Circuit Evaluation

Short-Circuit evaluation enables you to not evaluate the right-hand side of AND and OR expressions, when the overall result can be predicted from the left-side value.

int numberOne = 1;
int numberTwo = 2;
boolean result = false;

// left-side is false so the the overall result CAN be predicted without evaluating the right side.
// numberOne will be 1, numberTwo will be 2, result will be false
result = (numberOne > numberTwo) && (++numberOne == numberTwo);

System.out.println(numberOne); // prints 1
System.out.println(numberTwo); // prints 2
System.out.println(result);    // prints false


// left-side is true so the the overall result CAN NOT be predicted without evaluating the right side.
// numberOne will be 2, numberTwo will be 2, result will be true
result = (numberTwo > numberOne) && (++numberOne == numberTwo);

System.out.println(numberOne); // prints 2
System.out.println(numberTwo); // prints 2
System.out.println(result);    // prints true

& | ^ Full Evaluation

Although in some cases it is possible to predict the result, It is necessary to evaluate the right-hand side.

int numberOne = 1;
int numberTwo = 2;
boolean result = false;

// left-side is false so the the overall result will be false BUT the right side MUST be evaluated too.
// numberOne will be 2, numberTwo will be 2, result will be false
result = (numberOne > numberTwo) & (++numberOne == numberTwo);

System.out.println(numberOne); // prints 2
System.out.println(numberTwo); // prints 2
System.out.println(result);    // prints false

Notice:

1. Notice that for XOR (^) there is no short-circuit, because both sides are always required to determine the overall result.
2. Notice that other possible names for Short-Circuit evaluation are minimal evaluation and McCarthy evaluation.
3. It is not recommenced to mix boolean logic and actions in the same expression
4. & can also act as a Bitwise AND operator which is very academic and can be used in cryptography. When both bits are 1, the result is 1, or either of the bits is not 1, the result is 0. (Check the following code)

AND Bitwise example:

byte a = 5;              // 00000101
byte b = 3;              // 00000011
byte c = (byte) (a & b); // 00000001 (c is 1)

Answer 15

Almost every point of comparison is very well covered in all the answers. I just want to add one example. To demonstrate how the output changes based on which operator we use. Consider the below example

int a = 10;
if(++a==10 & ++a==12) {
    ++a;
}
System.out.println(a); //12

In the above code, We are using bitwise & operator. So It will evaluate both the arguments(left and right) irrespective of the individual result.

so a will increment 2 times within if condition. But as the condition will not become true, It will not enter inside the if-loop and 3rd increment will not happen. So the final value of a would become 12 in this case.

Now suppose, in the same above example If we use short-circuit && operator. then after evaluating ++a==10 to false, It will not go to check the second argument. And Hence the final value of a would-be 11.

int a = 10;
if(++a==10 && ++a==12) {
    ++a;
}
System.out.println(a); //11

Based on this, We can say that performance of bitwise & operator is relatively low compare to the short-circuit && operator. As bitwise operator will go to evaluate both the arguments irrespective of the result of the first argument. While && operator will stop evaluating the second argument if the first argument's result is false.

One more difference between these two is, Bitwise & operator is applicable for boolean as well as integral types. While short-circuit && operator is applicable only for the boolean type.

We can write

System.out.println(4 & 5); // 4

But if We try to write like ,

System.out.println(4 && 5);

Then it will give an error saying,

bad operand types for binary operator '&&'