Wrong results while String comparison

Question

I'm in little trouble. The problem is when I'm trying to compare 2 strings(type String) operator '==' returns FALSE, but actually strings are equal. Here's the code with its problem:

//before the following code I filled the "LinkedList <String> command" and there is
//a node with value of args[0]
String deal="";
Iterator it = commands.listIterator();
if(it.hasNext() == true)
{
    if(it.next() == args[0])
    {
        deal += it.next();
        it.hasNext();
        break;
    }
}

Thank You!!!

score 1 · Answer 1 · answered Sep 28 '14 at 20:38

1

You use .equals when comparing two strings. So use

(it.next()).equals(args[0])

answered Sep 28 '14 at 20:38

PsyCode

644
5
14

What a dummy I am. The problem is that I usually use C/C++ =) – Maxim Gotovchits Sep 28 '14 at 20:41

Alboz · Accepted Answer · 2014-09-28T20:43:51.293

1

To compare two strings u should use the method equals() or equalsIgnoreCase().

in your case:

if(it.next().equals(args[0]))

the operator == returns true if the two object are the same object, same address in memory.

edited Sep 28 '14 at 20:43

answered Sep 28 '14 at 20:38

Alboz

1,833
20
29

That wouldn't compile, you forgot the "." between next() and equals – PsyCode Sep 28 '14 at 20:39
Oooh yes! Thanks a lot. How could I forget that... Ofcourse our lecturer told us 'bout equals(...) – Maxim Gotovchits Sep 28 '14 at 20:40
@Maxim Gotovchits np, it's a common error when you start programming in java. – Alboz Sep 28 '14 at 20:43

score 1 · Answer 3 · answered Sep 28 '14 at 20:40

1

You have to use .equals method:

String deal="";
Iterator it = commands.listIterator();
if(it.hasNext() == true)
{
    String next = it.next();
    if(next.equals(args[0]))
    {
        deal += next;
        break;
    }
}

Be careful, .next() returns the value once and move its internal cursor to the next value.

The == cannot be used for String because the == is true if the same object instance is on both sides. The same string content can be in many String instances.

answered Sep 28 '14 at 20:40

Nicolas Albert

2,586
1
16
16

Thanks a lot. I have just forgotten that because I used to make programs on C/C++ =) – Maxim Gotovchits Sep 28 '14 at 20:45
Cool, you can vote up and accept my detailed answer :) – Nicolas Albert Sep 28 '14 at 20:46

score 1 · Answer 4 · answered Sep 29 '14 at 08:11

There are two ways of comparing strings.

Comparing the value of the strings (achieved using .equals ).
Comparing the actual object (achieved using == operator).

In your code you are comparing the references referred by it.next() & args[0]whereas you should compare the value of the two using it.next().equals(args[0]).

score 0 · Answer 5 · answered Sep 29 '14 at 11:43

if you use == to compare two int values, then it is compare the two values, because int is primitive data type. If you use "==" to compare String object, it is check whether both String reference are referring the same String object or not. It do not consider values of the String objects.

If you want to compare values of String objects you have to use equals() of the String class. This method is comparing content of both String objects.

Wrong results while String comparison

5 Answers5