CODEWITHSUNDEEP
javaoperator-keyword
Raju Boddupalli
Raju Boddupalli

Reputation: 1819

precedence of && and || operators in java

    Postfix operators [] . (parameters) expression++ expression--
Unary prefix operators ++expression --expression +expression -expression ~ !
Unary prefix creation and cast new (type)
Multiplicative * / %
Additive + -
Shift << >> >>>
Relational < <= > >= instanceof
Equality == !=
Bitwise/logical AND &
Bitwise/logical XOR ^
Bitwise/logical OR |
Conditional AND &&    //here
Conditional OR ||     //here   
Conditional ?:
Assignment = += -= *= /= %= <<= >>= >>>= &= ^= |=

This are the operator precedence s published in kalid A Mughal & Rasmussen book

according to this book && has higher precedence than || but the following code

  if(true&&false||true)
{
    System.out.println("yes");
}

this code prints "yes". means first executing "||"

is this book wrong? or my interpretation wrong.

Upvotes: 0

Views: 2395

Answers (8)

Jason Carter
Jason Carter

Reputation: 49

The && is evaluated first... but one side of the "or" || is true, so the overall expression evaluates to "true"

if(true&&false||true)
{
    System.out.println("yes");
}

Upvotes: 0

Elliott Hill
Elliott Hill

Reputation: 961

If you change your representation you can see it much clearer, for example:

if(true || true && false) {
    System.out.println("yes");
} else {
    System.out.println("no");
}

If && takes precedence you have ( true || (true && false) ) this will be true so will print "yes".

If || takes precence you will have ( (true || true) && false) this will be false and will print out "no.

If you run the above code you will see that the result is "yes" and therefore && has precendence.

Upvotes: 0

Eugene
Eugene

Reputation: 121068

look carefully, as these are short-circuit AND and OR

So the first part : 

 true&&false // evaluates to false

Then you have :

 false || true

the first part is false, BUT the second part could be true (and in fact it is), thus it moves to check the second part, sees that it's true, thus returns true

Upvotes: 0

Eng.Fouad
Eng.Fouad

Reputation: 117675

if(true && false || true) ==> if(false || true) ==> if(true).

AND Truth Table:

 F & F = F
 T & F = F
 F & T = F
 T & T = T

OR Truth Table:

 F | F = F
 T | F = T
 F | T = T
 T | T = T

Upvotes: 0

Affe
Affe

Reputation: 47994

Your interpretation is wrong

true && false -> false

false || true -> true

It is expected to print "yes"

Upvotes: 0

secretformula
secretformula

Reputation: 6432

  1. true && false => false
  2. false || true => true

Thus there is no issue as the and statement is executing first then the or statement. The expected result of the code is yes.

Upvotes: 0

kurtzbot
kurtzbot

Reputation: 512

Your interpretation is off. If we follow the book, your if-statement can be re-written and maintain the same meaning

if( (true && false) || true)

Now it is very easy to see that this does in fact evaluate to true, and therefore print out "yes".

Upvotes: 0

Tudor
Tudor

Reputation: 62469

I guess your interpretation is wrong. Since && has higher precedence the order of evaluation is

(true && false) || true

which is 

false || true -> true

Upvotes: 3

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