Creating Strings from Bytes/Ints in Java

Question

I'm wondering why the following code doesn't work:

String test = new String(new byte[] {92, 92, 92, 92, 92});
System.out.println(test);
String compare = "\\\\\\\\\\";
System.out.println(compare);
if (test == compare) {
System.out.println("Yes!");
}

The output is:

\\\\\
\\\\\

Where is a data type conversion happening that I'm not understanding?

Edit: /fail :(

Possible dup.: stackoverflow.com/questions/767372/java-string-equals-versus — Csaba_H
– Csaba_H, Commented Aug 27, 2009 at 16:57
its a dupe if you think i thought it was a problem with the equality i was using, yeah. if you read the title, i'm sure it will become obvious i had no idea this was a problem with equality — Tony Stark
– Tony Stark, Commented Aug 27, 2009 at 17:02

Uri · Accepted Answer · 2009-08-27 16:56:02Z

9

Strings are compared with .equals(), not with ==

The reason is that with references (as string variables are), == just checks equality in memory location, not in content.

The literal \\\ existed in one place in memory. the other one is created somewhere else where you build the string. They're not in the same location, so they don't return true when you do ==

You should do

if(test.equals(compare))

answered Aug 27, 2009 at 16:56

Uri

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1 Comment

captncraig Over a year ago

That is correct. You are doing a reference comarison instead of a value comparision. They are not the same object, but they are equivalent. Use Equals and is should work.

Calum · Accepted Answer · 2009-08-27 16:56:53Z

5

Strings in Java are reference types, and == checks whether they are the same string, rather than equal strings. Confusing I know. Long story short you need to do this:

if( test.equals( compare ) ) {

For more you can see here: http://leepoint.net/notes-java/data/strings/12stringcomparison.html

answered Aug 27, 2009 at 16:56

Calum

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Comments

sk. · Accepted Answer · 2009-08-27 16:57:26Z

3

You are testing to see if those are the same object, not if they are equal strings.

However the following test will be true:

test.intern() == compare.intern()

answered Aug 27, 2009 at 16:57

sk.

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Comments

erickson · Accepted Answer · 2009-08-27 16:56:39Z

2

You are using identity comparison, rather than string comparison.

Try test.equals(compare). Then try test.intern() == compare. Both should work. The intern method is the only reliable way to perform object identity comparisons on String objects.

answered Aug 27, 2009 at 16:56

erickson

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2 Comments

Calum Over a year ago

Just to raise a point about this answer, the "intern()" one only requires "test" to be intern()ed because "compare" is a literal in the source file (which means it's interned already). If it was created in the same way as "test", you'd need to use test.intern() == compare.intern() (and it's usually bad practice unless there's guaranteed to be a relatively small, or at least constrained, number of distinct strings).

erickson Over a year ago

Exactly! Thanks for pointing that out, Calum. Interning strings might be useful for an XML processor or something, where huge numbers of Strings could be created. It's sort of a Flyweight Pattern.

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