Boolean Functionality in Java - java

String s1="hi";
String s2="hi";
boolean b1 = true;
boolean b2 = false;
(1) System.out.println(s1==s2); //true
(2) System.out.println(s1==s2 + s1==s2); //false
(3) System.out.println(s1==s2+ " " + s1==s2); //false
(4) System.out.println(b1+b2); //error : bad operand types
(5) System.out.println(b1 + " " + b2); //true false
(6) System.out.println(true +" "+ s1==s2); //false
What is the difference between (2) & (4)?
What is the difference between (3) & (5)?
Why it gives result false in (3) & (6)?

Except for 4, all of these rely on operator precedence.
And in Java, + has precedence over ==.
Which means 2 actually "reads":
s1 == ((s2 + s1) == s2)
Therefore the right side operand of the first == is a boolean expression which compares two object references to one another (the fact that they are both Strings here is irrelevant) and here they are not the same. Hence the right side operand is boolean false.
But since the left side operand is a String, and since == is not applicable to operands String and boolean, this gives a compile error. JLS, section 15.21:
The equality operators may be used to compare two operands that are convertible (§5.1.8) to numeric type, or two operands of type boolean or Boolean, or two operands that are each of either reference type or the null type. All other cases result in a compile-time error.
If this really compiles for you, you are using a buggy Java compiler which autoboxes the right side operand to a Boolean, which it shouldn't. Let me guess: Eclipse's ECJ?
4 is an error since the + operator doesn't accept booleans as operands.
3 reads nearly the same as 2, except that this time it is s2 + " " + s1 which is (attempted to be) compared to s2. It fails to compile for the same reason.
In 5, booleans are autoboxed because of string concatenation.
6 again relies on the operator priority mentioned in 2; this time it is string true + " " + s1 which is (reference) compared with s2 (and that gives false). See 5 for what happens to true.

What is the difference between (2) & (4)?
Your second statement simply becomes System.out.println(hi == hihi ==hi); and the answer is false but your 4th statement is straight forward
What is the difference between (3) & (5)?
This is same as your previous question
Why it gives result false in (3) & (6)?
Your 3rd statement follows the above where as 6th statement convers as System.out.println("true hi" =="hi"); and output is false
P.S : '+' operator comes first in operator precedense

1) What is the difference between (2) & (4) ?
Ans :- == has more precedence over + so 2nd is actually reads as a " s1 == ((s2 + s1) == s2) " where we can't even imagine about ( boolean+boolean )
2) What is the difference between (3) & (5)?
Ans:- 3rd won't even compile and 5th is simple put values b1=true and b2=false and simple String concatenation performed

Related

Java: == in print statement gives different answers [duplicate]

This question already has answers here:
String equals and == with String concatenation [duplicate]
(4 answers)
Closed 7 years ago.
String a = "abc";
String b = "abc";
System.out.println("Result .... " + a==b); // false
System.out.println(a==b); // true
1st print statement prints false and 2nd prints true, though ideally it has to be true. Why is it false in 1st print statement ?
System.out.println("Result .... " +a==b); -> the result string will be appended with 'a' and then compares with b so it results false.
Order of operations:
"Result .... " + a==b
is equivalent to
("Result .... " +a) == b
which will be false since the two strings are not the same reference.
The explanation for this is that the + addition operator has a higher precedence than == logical equivalence.
The expression a == b is returning true in your second statement due to interning, in which a and b actually refer to same string object.
Click here for a link to Oracle's table of operator precedence in Java.
Forgot checking equality by == in java. In Java this operation checks equality of object link. Additionally it is applicable for checking equality of simple numbers. You should use .equals method
String a = "abc";
String b = new String(a);
System.out.printLn(a == b);//false
System.out.println(a.equals(b));//true
Learn about operation order in java
It is because in "Result .... " +a==b it first add "Result .... " with a and then == to b. If you write like this "Result .... " +(a==b), then it will be OK.
In the first statement, you're comparing "Result .... " + a with b. In the second one, you're comparing a with b, hence the difference. Change your first statement as follows:
System.out.println("Result .... " + (a==b));
And keep in mind that strings should be compared using the equals() method instead of ==.
System.out.println("Result .... " +a==b);
String Result is appended with 'a' and then compares with b so it provides false. (Result .... a == b) which is false.
Follow this link to understand precedence of operators in java and this.
Operator + has more precedence than == operator.
try adding brackets, you will see the diff. Bracket is evaluated separately.
public static void main(String[] args) {
String a = "abc";
String b = "abc";
System.out.println("Result .... " + (a == b)); // false
System.out.println(a == b); // true}
}
output
Result .... true
true

Why should I always use || instead of | and && instead of &?

Since I started programming Java, I've noticed that everyone was using && and || instead of & and |. What is the reason for this? I've been using && and || all this time because I didn't know you can use & and | on booleans.
class A{static{
boolean t = true;
boolean f = false;
System.out.println("&ff " + (f&&f) + " " + (f&f));
System.out.println("&ft " + (f&&t) + " " + (f&t));
System.out.println("&tf " + (t&&f) + " " + (t&f));
System.out.println("&tt " + (t&&t) + " " + (t&t));
System.out.println("|ff " + (f||f) + " " + (f|f));
System.out.println("|ft " + (f||t) + " " + (f|t));
System.out.println("|tf " + (t||f) + " " + (t|f));
System.out.println("|tt " + (t||t) + " " + (t|t));
}}
As far as I can tell, they are the same:
$ javac A.java && java A
&ff false false
&ft false false
&tf false false
&tt true true
|ff false false
|ft true true
|tf true true
|tt true true
Exception in thread "main" java.lang.NoSuchMethodError: main
Does using || and && improve my code in any way?
As a simple test, I replaced hundreds of occurrences with the short form, and all of my unit tests still pass.
|| and && uses short circuit evaluation
From same article
Short-circuit evaluation, minimal evaluation, or McCarthy evaluation
denotes the semantics of some Boolean operators in some programming
languages in which the second argument is only executed or evaluated
if the first argument does not suffice to determine the value of the
expression
Consider you have an object and you want to check one of its property for some value if you do:
if(obj != null & obj.ID == 1)
If your object obj is null you will get a null reference exception, since you used single &, the first condition evaluate to false, but it will still continue to the next condition and raising null reference exception.
If you used && then the second condition will never get evaluated, thus no exception.
if(obj != null && obj.ID == 1)
With your code you will not see any difference, but using bitwise | or & with multiple conditions would result in no short circuiting at all.
More Explanation.
consider that you following code:
Boolean conditionA = 2 > 1; //some condition which returns true
if(conditionA | obj.SomeTimeConsumingMethod()){
}
Now in above code snippet, suppose you have an object with a method SomeTimeConsumingMethod which takes a lot of time in processing and returns true or false. If you use single | it would evaluate both the conditions, since first conditionA is true it will process obj.SomeTimeConsumingMethod as well. End result will be true for the whole if statement since | (OR) is used.
If your condition is using double || (OR Logical operator)
if(conditionA || obj.SomeTimeConsumingMethod())
Then the second condition obj.SomeTimeConsumingMethod() will not be evaluated. This is short circuiting and that just saved you from executing some time consuming method. Still the end result is true regardless of what was returned from obj.SomeTimeConsumingMethod().
As has already been pointed out, && and || do short-circuit evaluation. & and |, with boolean operands, do the same operations but evaluate both operands regardless of the value of the first operand.
There are cases where it matters which you use, such as when the left operand is a pre-condition for exception-free evaluation of the right hand operand.
In most cases, it does not matter. I use && and || in those cases for readability, because that is the commoner choice. Using & or | with boolean operands would tend to make most Java programmers have to stop and think, and ask themselves whether there is some specific reason for it. I suspect it may be so common because of the inheritance from C and C++.
|| is logical or, where | is the bitwise operation or. Same with && and &. Use && and || if you're in an if statement, and | or & if you're doing bit operations.
While you may get the "correct" result using the bit-wise operators, understand they do NOT represent the same thing. For instance, this works
public static void main(String[] args){
int x = 5;
int y = 9;
System.out.println(x&y); // prints "1"
}
However, this won't
public static void main(String[] args){
int x = 5;
int y = 9;
System.out.println(x&&y); // Compile error: java: operator && cannot
// be applied to int,int
}
The reason you don't use bit-wise operators for boolean logic is so that it's clear what the code is trying to do (as well as knowing that it actually works). Bit-wise operators are for manipulating bit values, where as boolean operators are for evaluating first-order logic statements.
While not the same, it is along the same reasoning for using .equals() vs == for comparison -- You don't want readers guessing as to what you meant to do. It may work in some instances (like comparing constant, hard-coded strings with ==), but it is poor form because what it says is that you are asking for instance equality as opposed to value equality, which is usually implied. And using bit-wise operators implies you want to do bit manipulation, not first-order logic evaluation.
Arguing that short circuit evaluation is mandatory because of performance is weak. You can make an argument that is almost as weak: that short circuit evaluation encourages extra nondeterminism and side channels. Random optimizations or random loss of nondeterminism/side channels, I'd prefer the latter.
The only remaining argument is this: If your code base relies on short circuit behaviour, e.g., the trick described by Habib (if (obj != null && obj.ID == 1)), then it may not be a good idea to start mixing in &, because programmers might get it confused with &&. i.e., if you use &/| everywhere, but only use && in a few places where you rely on short circuit evaluation, someone might make an error and put only &, it may be hard to spot. In this case, if you relied on && for performance instead of something like null checking, it may be harder to spot when someone accidentally puts & alone, since there would be no stacktrace.
Does using || and && improve my code in any way?
Let me demonstrate two examples on how the conditional boolean operators may improve your code. Compare
if (a != null && a.equals("b") || z != null && z.equals("y"))
System.out.println("correct");
with
boolean b1 = a != null, b2 = z != null;
if (b1) b1 = b1 & a.equals("b");
if (b2) b2 = b2 & z.equals("y");
if (b1 | b2) System.out.println("correct");
which is the best way I can think of relying solely on the logical boolean operators.
Second, say you have two tests to perform: simpleTest() and heavyTest(), where the latter involves making an HTTP request and parsing the response. You have
if (simpleTest() & heavyTest()) System.out.println("success");
else throw new IllegalStateException();
and want to optimize it. What would you rather do, this:
if (simpleTest()) {
if (heavyTest()) System.out.println("success");
else throw new IllegalStateException("operation could not be completed");
} else throw new IllegalStateException("operation could not be completed");
this:
boolean b = simpleTest();
if (b) b = b & heavyTest());
if (b) System.out.println("success");
else throw new IllegalStateException("operation could not be completed");
or this:
if (simpleTest() && heavyTest()) System.out.println("success");
else throw new IllegalStateException();
Why should I always use || instead of | and && instead of &?
You shouldn't always use them, but you would do yourself a favor by making them your default because in 99.9% of uses they either don't hurt or make your code much better. Reserve the logical operators for that 0.1% of cases where you truly do need their semantics.
It has been mentioned elsewhere on this page that relying on the conditional evaluation of the right operand makes for counterintuitive code. I refuse to accept such a view on intuition because, by definition, intuition is a learned skill. It is a facility bestowed to you by evolution to recognize a pattern in situations you encounter often, and to quickly (and subconciously) cut through to the correct conclusions. One of the most important aspects of learning a language is acquiring the right intuitions. The argument that a language feature is bad because it is counterintuitive to someone who looks at it for the first time cannot be taken seriously.
In Java (and C/C++ if I remember correctly) the single "|" and "&" are bit-wise operators, not logical operators, they are used for completely different things.
You use the "&&" and "||" for boolean statements
You use the "&" and "|" for bit operations, given the following are binary numbers
dont make assumption so fast. It because this wont compile:
static boolean f(boolean a) {
boolean x = (true|a) ? a : x;
return x;
}
Bro.java:6: variable x might not have been initialized
boolean x = (true|a) ? a : x;
^
and this will:
static boolean f(boolean a) {
boolean x = (true||a) ? a : x;
return x;
}
&& and || are short cut operators. For example:
A && B
Condition B will not be evaluated if condition A is false
A || B
Condition B will not be evaluated if condition A is true
You should always use shortcut operators for multiple conditions evaluation. | or & are used for bit-wise operations.

Java: boolean in println (boolean ? "print true":"print false") [duplicate]

This question already has answers here:
What is the Java ?: operator called and what does it do?
(17 answers)
Closed 9 years ago.
I came across this syntax:
System.out.println(boolean_variable ? "print true": "print false");
What is this syntax with two dots : called?
Where can I find info about it?
Does it work just for booleans or is it implemented in other different ways?
? : is the conditional operator. (It's not just the : part - the whole of the method argument is one usage of the conditional operator in your example.)
It's often called the ternary operator, but that's just an aspect of its nature - having three operands - rather than its name. If another ternary operator is ever introduced into Java, the term will become ambiguous. It's called the conditional operator because it has a condition (the first operand) which then determines which of the other two operands is evaluated.
The first operand is evaluated, and then either the second or the third operand is evaluated based on whether the first operand is true or false... and that ends up as the result of the operator.
So something like this:
int x = condition() ? result1() : result2();
is roughly equivalent to:
int x;
if (condition()) {
x = result1();
} else {
x = result2();
}
It's important that it doesn't evaluate the other operand. So for example, this is fine:
String text = getSomeStringReferenceWhichMightBeNull();
int usefulCharacters = text == null ? 0 : text.length();
It's the conditional operator, often called ternary operator because it has 3 operands: An example would be:
int foo = 10;
int bar = foo > 5 ? 1 : 2; // will be 1
int baz = foo > 15 ? 3 : 4; // will be 4
So, if the boolean expression evaluates to true, it will return the first value (before the colon), else the second value (after the colon).
You can read the specifics in the Java Language Specification, Chapter 15.25 Conditional Operator ?
It's a ternary operator, meaning that instead of having two operands like many other operators, it has three. Wikipedia on Ternary Operation and how it's used in Java. What it boils down to: the boolean operation (or just a variable) is evaluated. If it evaluates to true, the operator returns the value / executes the code before the :, otherwise the one after it.
That's an if statement.
What's to the left of ? is the condition, what's between the ? and : is the result if the condition is true, and what's to the right of : is the result if the condition is false.
This is ternary operator (http://en.wikipedia.org/wiki/?:). It can be used anywhere when you need a small if expression.
For your questions:
The ?: (both characters together) are called conditional operator (or ternary operator). Only both together will work.
Search for java ternery operator
It only works for boolean
In principle the ternery operator is a shortened if/else. The boolean will be the condition to the if, the part between ? and : is the if branch and the part after this is the else branch.
Please note that the return type of the conditional operator is determined by the first branch.
It's the ternary operator and it works with booleans. It can be used as a shorthand for if-else in some cases, but shouldn't be used for too complicated things as it can be difficult to read.
An example would be assigning value to a variable depending on a condition:
String message = doOperation() ? "Success" : "Error occurred";
System.out.println(message);
In this case, if doOperation returns a boolean telling whether it succeeded or not, the message to be shown can be assigned on a single line.
Please note that this example does not represent good programming practices.
Its ternary operator.
The ternary operator or ?, is a shorthand if else statement. It can be used to evaluate an expression and return one of two operands depending on the result of the expression.
boolean b = true;
String s = ( b == true ? "True" : "False" );
This will set the value of the String s according to the value of the boolean b. This could be written using an if else statement like this:
boolean b = true;
String s;
if(b == true){
s = "True";
}else{
s = "False";
}
Its a short form of if-else statement.
It works in this way
(yourCondition ? STATEMENT1 : STATEMENT2)
The compiler checks for the condition.
IF it returns TRUE then STATEMENT1 will be executed.
ELSE STATEMENT2 will be executed.
The question mark followed by a colon (two dots) is a ternary operator usually called inline if.
In this case it returns a string depending on the value of boolean_variable.
http://en.wikipedia.org/wiki/%3F:
See here. The ternary operator is similar to an if expression but differs in that it is an expression - it has a return value, while if expressions don't. Sometimes you want to use it to make your code a little less cluttered.

What does "?" mean in Java? [duplicate]

This question already has answers here:
What is the Java ?: operator called and what does it do?
(17 answers)
Closed 10 years ago.
I dont know what the question mark (?) stand for in java, I was doing a small program, a Nim-game. were looking in a book, for help and saw this statement:
int pinsToTake = (min >= 2) ? 2 : 1;
I don't understand it, what will ? represent, can it be something to do with if-statement but you put it in a variable? and the : can be something "else"? (this things that I just said can be very misleading)
someval = (min >= 2) ? 2 : 1;
This is called ternary operator, which can be used as if-else. this is equivalent to
if((min >= 2) {
someval =2;
} else {
someval =1
}
Follow this tutorial for more info and usage.
Its ternary operator also referred to as the conditional operator, have a look reference
like Object bar = foo.isSelected() ? getSelected(foo) : getSelected(baz);
eg. operand1 ? operand2 : operand3
if operand1 is true, operand2 is returned, else operand3 is returned
operand1 must be a boolean type
operand1 can be an expression that evaluates to a boolean type
operand1 and operand2 must be promotable numeric types or castable object references, or null
if one of operand2 or operand3 is a byte and the other a short, the type of the returned value will be a short
if one of operand2 or operand3 is a byte, short or char and the other is a constant int value which will fit within the other operands
range, the type of the returned value will be the type of the other
operand
otherwise, normal binary numeric promotion applies
if one of operand2 or operand3 is a null, the type of the return will be the type of the other operand
if both operand2 and operand3 are different types, one of them must be compatible (castable) to the other type
reference
it means:
if(min >= 2)
someval =2;
else
someval =1
Its called a ternary operator
See this java example too
That's a Ternary Operator. Check Oracle's doc for further info. Long story short, it is an if-else statement that can be done in a single line and used inside methods and to define variable values.
Syntax:
boolean_expression ? do_if_true : do_if_false;
Parallelism with if-else statement:
if(boolean_expression)
//do_if_true;
else
//do_if_false;
I didn't use brackets on purpose, since you can only execute one line of code in do_if_true and do_if_false.
Example of use:
boolean hello = true;
String greetings = hello ? "Hello World!" : "No hello for you...";
This will set someString as "Hello World!" since the boolean variable hello evaluates to true. On the other hand, you can nest this expressions:
boolean hello = true;
boolean world = false;
String greetings = hello ? (world ? "Hello World!" : "Hello Stranger!") : "No hello for you...";
In this case, greetings will have as a value "Hello Stranger!";
It's called the Ternary If operator, it's just short-hand for an if...else
"? :" is a ternary operator equivalent to an if else statement.
In your example:
pinsToTake = (min >= 2) ? 2 : 1
if min >= 2 then assign 2 to pinsToTake, else assign 1
max = (a > b) ? a : b;
(a > b) ? a : b; is an expression which returns one of two values, a or b. The condition, (a > b), is tested. If it is true the first value, a, is returned. If it is false, the second value, b, is returned. Whichever value is returned is dependent on the conditional test, a > b. The condition can be any expression which returns a boolean value.
It is called conditional operator.This is how it works.
if min is greater than or equal to 2 ,then first value after ? that is 2 here will be assign to corresponding variable ,otherwise second value that is 1 here will be assign.
This link will tell you all you need.
Summary for archival sakes:
It's called the conditional operator. It's a ternary operator that
takes three terms:
BooleanExpression ? Expr1 : Expr2
The BooleanExpressionis evaluated. If it's true, the value of the
whole expression is Expr1. If it's false, the value of the whole
expression is Expr2.
So it serves the same kind of purpose as an if statement, but it's a
term rather than a whole statement. That means you can embed it in
places where you can't use a whole statement.

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

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?
& <-- 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 ||.
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
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
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.
&& 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
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 any 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
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.
‘&&’ : - 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
&& 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.
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.
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
& 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.
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
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:
Notice that for XOR (^) there is no short-circuit, because both sides are always required to determine the overall result.
Notice that other possible names for Short-Circuit evaluation are minimal evaluation and McCarthy evaluation.
It is not recommenced to mix boolean logic and actions in the same expression
& 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)
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 '&&'

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