Integer to String conversion?

Question

I have a problem with Integer.toString conversion. This code outputs "ololo". Why? And how can I convert integer to string right?

 String str1= "1";
 String str2=Integer.toString(1);
 if (str1!=str2)Log.d("myLogs","ololo");

You must compare `String`s using `equals` method, not `==` nor `!=`. — Luiggi Mendoza, Apr 21 '13 at 14:50
Search for `String comparison Java`, you will get a lot of hits. — Keppil, Apr 21 '13 at 14:51
Why do you take toString() of the Integer?? Just use `String str2= 1 + "";` — Praneeth Peiris, Apr 21 '13 at 14:55
Or if you think it's not a good practice, use `String.valueOf(1);` And do NOT compare Strings with == or !=. Use `if(string1.equals(string2)){}` — Praneeth Peiris, Apr 21 '13 at 14:57
@GnomezGrave: There is absolutely nothing wrong with using `Integer.toString()`. — Keppil, Apr 21 '13 at 15:00
I meant, the other way is the easy way.. or the lazy way. :) — Praneeth Peiris, Apr 21 '13 at 15:13

score 2 · Answer 1 · answered Apr 21 '13 at 14:51

You must compare Strings using equals method, not == nor != operators since they will compare the String object references.

if (!str1.equals(str2)) {
    Log.d("myLogs","ololo");
}

Note that when you use Integer#toString you're creating a new String that is not in the String JVM pool, thus getting the error described.

score 0 · Answer 2 · answered Apr 21 '13 at 14:52

0

String comparison must be done with equals.
if (!str1.equals(str2))...

When you use != you get reference equality (inequality)

answered Apr 21 '13 at 14:52

Itay Karo

Achintya Jha · Answer 3 · 2013-04-21T15:00:49.200

0

Try String.valueOf(1); to change Integer to String.

edited Apr 21 '13 at 15:00

answered Apr 21 '13 at 14:52

Achintya Jha

score 0 · Answer 4 · edited Apr 21 '13 at 14:57

0

use !str1.equals(str2) instead.

You shouldn't use == or != for String

edited Apr 21 '13 at 14:57

nKandel

answered Apr 21 '13 at 14:52

Nicholas Kadaeux

4 Answers4