CODEWITHSUNDEEP
javaboolean
emsnguyen
emsnguyen

Reputation: 41

How is (true != false != true) different from (false != true != false)?

I recently took an entry test in Java and this question confused me. The full question is:

boolean b1 = true;
boolean b2 = false;
if (b2 != b1 != b2) 
    System.out.println("true");
else 
    System.out.println("false");

My first question is what (b2 != b1 != b2) means and the second question, as specified in the title, is how (false != true != false) evaluates to true while (true != false != true) evaluates to false (I tested that on Netbeans).

Upvotes: 2

Views: 1196

Answers (3)

davidxxx
davidxxx

Reputation: 131386

You have two boolean comparisons where the first comparison produces a result that is compared to another boolean value (the last one).
And the equality operators are syntactically left-associative (they group left-to-right).

To understand you can rewrite the actual comparison by doing the comparison in two times :

1) false != true != false == true as 

boolean result = false != true; // ->true
result = true != false; // ->true
result == true;

2) true != false != true == false as 

boolean result = true != false; // -> true
result = true != true; // -> false
result == false;

Or you can also enclose the fist comparison by parenthesis to ease the reading of the evaluations precedence (left to right) :

1) false != true != false == true as 

 <=> (false != true) != false 
 <=>      true       != false
 <=>      true

2) true != false != true == false as 

 <=> (true != false) != true 
 <=>     true        != true
 <=>     false

Upvotes: 6

Raman Mishra
Raman Mishra

Reputation: 2686

See what happens most of the compiler start evaluating the expression from left to right So in this case what is happening first it is evaluating this.

False != True == True //is evaluated first which is true

Then it evaluates with this true we got from the first expression

True != False == True //which is also true

So the completer expression goes like this

False != True != False == True// which is True

Now in the second case the expression is like

True != False != True == False

the outputs for fist expression which is True != False evaluates for True and expression becomes True != True which is false

I hope it makes sense. It’s because the associativity of != is from left to right so it evaluating from left to right.

Upvotes: 0

Nutan
Nutan

Reputation: 786

it will evaluate this way :

1-false != true != false = false!=true=true

so this false != true became true and then equation become like this

true != false which is equal to true

so the result is true.

Now for second one you can evaluate the same way as

2- true != false != true

true!=false which is true

now true != true which is false so you are getting result as false

Upvotes: 0

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