Java String.equals versus ==

Question

This code separates a string into tokens and stores them in an array of strings, and then compares a variable with the first home ... why isn't it working?

public static void main (String... aArguments) throws IOException {

String usuario = "Jorman";
String password = "14988611";

String strDatos="Jorman 14988611";
StringTokenizer tokens=new StringTokenizer(strDatos, " ");
int nDatos=tokens.countTokens();
String[] datos=new String[nDatos];
int i=0;

while(tokens.hasMoreTokens()) {
    String str=tokens.nextToken();
    datos[i]= str;            
    i++;
}

//System.out.println (usuario);

if( (datos[0]==usuario)) {  
     System.out.println ("WORKING");    
}

Answer 1

Use the String.equals(String other) function to compare strings, not the == operator.

The function checks the actual contents of the string, the == operator checks whether the references to the objects are equal.

if (usuario.equals(datos[0])) {
    ...
}

nb: the compare is done on 'usuario' because that's guaranteed non-null in your code, although you should still check that you've actually got some tokens in the datos array otherwise you'll get an array-out-of-bounds exception.

Answer 2

Instead of

datos[0] == usuario

use

datos[0].equals(usuario)

== compares the reference of the variable where .equals() compares the values which is what you want.

Answer 3

You should use string equals to compare two strings for equality, not operator == which just compares the references.

Answer 4

It's good to notice that in some cases use of "==" operator can lead to the expected result, because the way how JVM handling strings - string literals are interned (see String.intern()) - so when you write for example "hello world" in two classes and compare those strings with "==" you could get result: true, which is expected, but it's only side effect of jvm optimizations; when you compare same strings (linguistically) when the first is string literal (ie. defined through "string") and second is constructed through "new" keyword like new String("string") using of == (equality operator) returns false, because both of them are different instances of String class.

Only right way is using ".equals()" -> datos[0].equals(usuario). "==" says only if two objects are the same instance of object (ie. have same memory address)

Answer 5

For anyone who needs more proves:

String abcdString = "abcd";
char[] abcdArray = {'a','b','c','d'};
String abcdArrayToString = new String(abcdArray);
System.out.println("abcdString: "+abcdString);
System.out.println("abcdArrayToString: "+abcdArrayToString);
System.out.println("abcdString == abcdArrayToString = "+(abcdString == abcdArrayToString));
System.out.println("abcdString.equals(abcdArrayToString) = "+abcdString.equals(abcdArrayToString));

Answer 6

It will also work if you call intern() on the string before inserting it into the array. Interned strings are reference-equal (==) if and only if the are value-equal (equals().)

public static void main (String... aArguments) throws IOException {

String usuario = "Jorman";
String password = "14988611";

String strDatos="Jorman 14988611";
StringTokenizer tokens=new StringTokenizer(strDatos, " ");
int nDatos=tokens.countTokens();
String[] datos=new String[nDatos];
int i=0;

while(tokens.hasMoreTokens()) {
    String str=tokens.nextToken();
    datos[i]= str.intern();            
    i++;
}

//System.out.println (usuario);

if(datos[0]==usuario) {  
     System.out.println ("WORKING");    
}