Java: == in print statement gives different answers

Question

String a = "abc";
String b = "abc";

System.out.println("Result .... " + a==b); // false
System.out.println(a==b);                  // true

1^st print statement prints false and 2^nd prints true, though ideally it has to be true. Why is it false in 1^st print statement ?

@John3136 I know that very well. Better understand the question scenario. — ShruR, Jan 18 '16 at 05:44
Good stuff. Just as soon as I see `string == string` the alarm bells start to go off... — John3136, Jan 18 '16 at 05:46
Why ask same question: http://stackoverflow.com/questions/34848089/java-print-statement-only-prints-the-result? Are posted answers not clear? Could you explain what other informations you want to get? — Pshemo, Jan 18 '16 at 06:19

score 4 · Answer 1 · answered Jan 18 '16 at 05:41

4

System.out.println("Result .... " +a==b); -> the result string will be appended with 'a' and then compares with b so it results false.

answered Jan 18 '16 at 05:41

Shriram

4,343
8
37
64

score 4 · Answer 2 · edited May 23 '17 at 11:45

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.

score 2 · Answer 3 · edited Jan 18 '16 at 06:24

2

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

edited Jan 18 '16 at 06:24

Pshemo

122,468
25
185
269

answered Jan 18 '16 at 05:53

Pavel Uvarov

1,090
1
10
16

score 0 · Answer 4 · answered Jan 18 '16 at 05:43

0

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.

answered Jan 18 '16 at 05:43

Bahramdun Adil

5,907
7
35
68

score 0 · Answer 5 · answered Jan 18 '16 at 05:43

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 ==.

Ankur Singhal · Answer 6 · 2016-01-18T05:58:51.953

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

Java: == in print statement gives different answers

6 Answers6