strange behavior of == in Java

Question

I observed a strange behavior == operator in java. I am trying to print the out put as follows

String str1 = "Rajesh";
String str2 = "Rajesh";
System.out.println("Using equals() str1 and str2 Equals :"
            + str1.equals(str2));
System.out.println("Using == str1 and str2 Equals :" 
            + str1 == str2);

The first SOP statement printing

Using equals() str1 and str2 Equals :true

and the next SOP printing only false .

I tried compiling in both eclipse and Net Beans but result is the same . I am so confused why

Using == str1 and str2 Equals :

is not printing

Help me out in this

Thanks in advance,

Raj

Did you try encapsulating the condition in parenthesis? It's possible the first condition is added to your string then evaluated with str2. — Josh, Feb 09 '12 at 06:17
`"Using == str1 and str2 Equals :" + str1 == str2` will be the equivalent of `"Using == str1 and str2 Equals :" + "Rajesh" == "Rajesh"`. Because `+` has a higher precedence than `==` (`2 + 2 == 4`), we'll have the equivalent of `"Using == str1 and str2 Equals :Rajesh" == "Rajesh"`, which will be `false`. — Tom Hawtin - tackline, Feb 09 '12 at 06:20
possible duplicate of [Java String.equals versus ==](http://stackoverflow.com/questions/767372/java-string-equals-versus) — Michael Petrotta, Feb 09 '12 at 06:21
When you don't use any parentheses, the `==` operator is evaluated last. — Sam Dufel, Feb 09 '12 at 06:22

score 12 · Accepted Answer · answered Feb 09 '12 at 06:18

12

it's the same as ("Using == str1 and str2 Equals :" + str1) == str2 and this is false, of course. Expression is parsed from left to right and so at first it concatenates "Using == str1 and str2 Equals :" and str1, then applies == operator.

answered Feb 09 '12 at 06:18

shift66

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score 7 · Answer 2 · answered Feb 09 '12 at 06:26

See http://bmanolov.free.fr/javaoperators.php for a table of operator precedence in Java.

The + operator is higher precedence than the == operator.

So, in effect, your code is equivalent to the following:

System.out.println( ("Using == str1 and str2 Equals :" + str1) == str2);

Note the placement of the parentheses that I added. It evaluates to this:

System.out.println( (str_x + str1) == str2);

And then to this:

System.out.println( str_y == str2 );

And then to this:

System.out.println( false );

In order to get the result you want, you must use parentheses to specify that you want the == operator to be resolved BEFORE the + operator:

System.out.println( "Using == str1 and str2 Equals :" + (str1 == str2));

Notice the new placement of the parentheses.

score 4 · Answer 3 · answered Feb 09 '12 at 07:10

Because + has higher priority compare to = and if you use bracket(str1 == str2) then this result give true because highest priority is (. So First it checks bracket inside data.

String str1 = "Rajesh";
        String str2 = "Rajesh";
        System.out.println("Using equals() str1 and str2 Equals :"
                + str1.equals(str2));
        System.out.println("Using == str1 and str2 Equals :" 
                + (str1 == str2));

Output:

Using equals() str1 and str2 Equals :true
Using == str1 and str2 Equals :true

score 2 · Answer 4 · answered Feb 09 '12 at 06:17

2

Maybe an order of operations thing? Try:

System.out.println("Using == str1 and str2 Equals :" + (str1 == str2));

answered Feb 09 '12 at 06:17

Oleksi

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score 2 · Answer 5 · answered Feb 09 '12 at 06:18

2

Try surrounding it with () like this:

System.out.println("Using == str1 and str2 Equals :" + (str1 == str2));

answered Feb 09 '12 at 06:18

Kuldeep Jain

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It works fine and prints this: `Using == str1 and str2 Equals :true`. The explain is given by @Ademiban – Kuldeep Jain Feb 09 '12 at 06:33

score 0 · Answer 6 · answered Feb 09 '12 at 06:57

equals method returns true if and only if x and y refer to the same object.Follwoing is the Object class implementation of equals method.

public boolean equals(Object obj) {
    return (this == obj);
    }

In String class this method has overridden as following.

public boolean equals(Object anObject) {
    if (this == anObject) {
        return true;
    }
    if (anObject instanceof String) {
        String anotherString = (String)anObject;
        int n = count;
        if (n == anotherString.count) {
        char v1[] = value;
        char v2[] = anotherString.value;
        int i = offset;
        int j = anotherString.offset;
        while (n-- != 0) {
            if (v1[i++] != v2[j++])
            return false;
        }
        return true;
        }
    }
    return false;
    }

And if you use == operator it just check both references are having same object. Similar to the Object class equals method.

score -1 · Answer 7 · answered Feb 09 '12 at 06:18

-1

The reason is that you cannot compare strings in Java using ==.

In C++ or C# (or other languages supporting operator redefinition), you can overwrite the == operator to provide that functionality. Java does not support that.

answered Feb 09 '12 at 06:18

Abrixas2

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You mean overload, not overwrite (or override). And yes, you *can* compare string references using `==` in Java. It will compare the references, not the objects, but that would work fine in this particular case, if it had been used correctly. – Jon Skeet Feb 09 '12 at 06:25
Yes, I meant overload. And thanks for pointing on that reference thing. I just did not know that. – Abrixas2 Feb 09 '12 at 06:28

score -1 · Answer 8 · answered Feb 09 '12 at 06:22

-1

str1.equals(str2) returns true because the equals() function compares the content of the string variables, where as == operator compares the instances. Since str1 and str2 are two differences of instances of String class, it returns false

answered Feb 09 '12 at 06:22

Sunil Kumar B M

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Here, `str1` and `str2` will be identical references to the same string though, due to the way the language handles compile-time string constants. – Jon Skeet Feb 09 '12 at 06:33

score -1 · Answer 9 · answered Feb 09 '12 at 06:27

-1

In Java == operator matches the two objects i.e their address while .equals() method mathces the values of both objects, thats why you are getting true for equals() and false for == as both are different objects.

answered Feb 09 '12 at 06:27

Kamran Ali

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score -2 · Answer 10 · answered Feb 09 '12 at 06:17

-2

== can only be used to compare primitive datatypes. To compare objects you need to use equals method. Using a == operator on objects actually compares there addresses instead of values.

answered Feb 09 '12 at 06:17

Husain Basrawala

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Well, it compares the values - but those values are references. It's incorrect to say that `==` *can only be used to compare primitive datatypes* when it can also be used to test for reference equality. In this case, as we're dealing with two occurrences of the same string constant, `==` would work fine - if it were being used correctly. – Jon Skeet Feb 09 '12 at 06:23
Bearing in mind that the vast majority of objects out there do not override equals and simply return the result of ==. It's a fairly common mistake I see for people to just assume "I used .equals so this will work right." – Affe Feb 09 '12 at 06:23

strange behavior of == in Java

10 Answers10