CODEWITHSUNDEEP
javastringconcatenation
Monica
Monica

Reputation: 1599

Why am I getting different outputs with .concat() and += with Java Strings?

Studying for OCAJ7

I know String objects are immutable. I know using methods like .concat() on a String object will just create a new String object with the same reference name. I'm having a hard time, however, with wrapping my head around the following:

String str1 = "str1";
String str2 = "str2";
System.out.println( str1.concat(str2) );
System.out.println(str1);

// ouputs
// str1str2
// str1

String str3 = "fish";
str3 += "toad";
System.out.println(str3);

// outputs
// fishtoad

If strings are immutable, why does using += to concatenate affect the original String object, but .concat() does not? If I only want a String to concatenate with using +=, is using String better than using StringBuilder, or vice versa?

Upvotes: 2

Views: 130

Answers (7)

Devavrata
Devavrata

Reputation: 1785

Above you are not changing the reference of str1 in System.out.println(str1.concat(str2) ) so it is temporary storing the reference of str1.concat(str2).

But in str3+=str4 whole reference get moved to "fishtoad" therefore it get changed.But again if you defined a string str5="fish" then it reference back to previous location which was referencing by str3.

Upvotes: 0

Varun
Varun

Reputation: 1004

concat() method creates a new string. You have to remember that Strings in Java are immutable.

str1 will point to the newly created string if the o/p is assigned to it, 

str1 = str1.concat(str2);

otherwise str1 continues to point to the old string.

If you are going to concat multiple strings together in a loop, then it is better to use StringBuilder

Upvotes: 0

Nicholas Hollander
Nicholas Hollander

Reputation: 340

What's happening here is that String.concat(String); does not actually combine str1 and str2 together, instead it returns a new String object that is equivalent to str1str2. When you use str1 += str2, you are actually combining the variables, and then putting the value into str1.

Upvotes: 1

user63762453
user63762453

Reputation: 1734

Yes you are right Strings are immutable.

But when you write str1.concat(str2) inside System.out.println() you are printing the result returned by the concat() function.The result is a new String which is outputted on the console. You haven't assigned the value to str1.

But when you write += you are first concatenating something to the String the then assigning the reference back to str1.

This explains the output.

Upvotes: 1

rgettman
rgettman

Reputation: 178263

The concat method is more like + than +=. It returns the new string; it doesn't modify the original string or the argument (Strings are immutable).

str1.concat(str2)

is equivalent to str1 + str2, so str1 isn't modified, and the result is discarded.

However, += also assigns the result back to the left side, and str3 now refers to the new string. That's the difference.

Upvotes: 1

suman j
suman j

Reputation: 6960

str3 += "toad" means
str3 = str3 + "toad";

So you concatenate and assign the result back to str3

Upvotes: 0

Jigar Joshi
Jigar Joshi

Reputation: 240898

because you are catching the reference of newly generated String instance in str3

str3 += "toad";

is 

str3 = str3 + "toad"

Upvotes: 3

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