String concat

[jose chiramal]

String x="abc"
System.out.println(x.concat("def")); // output is abcdef


String x = "abc"
s.concat("def") // this object gets created but gets lost/abandoned , why so ?
System.out.println("x=" + x) // output is x=abc


Can someone explain why the above two print statements behave differently.

[Pushkar Choudhary]

jose chiramal wrote:String x = "abc"
s.concat("def") // this object gets created but gets lost/abandoned , why so ?
System.out.println("x=" + x) // output is x=abc

If you are doing s.concat(), obviously the value of x remains the same, i.e. "abc".

[Pushkar Choudhary]

Assuming the s.concat() is a typo, I think the reason for the output is that, after concatenation, you don't assign its value to anything.

jose chiramal wrote:this object gets created but gets lost/abandoned , why so ?

The object gets lost because you don't assign it to anything.

[Rob Spoor]

Strings are immutable, they never change. All methods that return a String return a different String object. The original String (s / x) stays the same even if you ignore the result of the concat method.

[jose chiramal]

In that case when I say

String x = "abc"

System.out.println(x.concat("def")) // output is x=abcdef

Does it mean that the System.out.println() by default calls a toString() that returns a new String Object ?

[roi yal]

man, String is final.