Are String.Equals() and == Operator Really Same

String.equals versus ==

Use the string.equals(Object 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. Note that string constants are usually "interned" such that two constants with the same value can actually be compared with ==, but it's better not to rely on that.

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.

String comparison : operator==() vs. Equals()

I like Equals() because the available StringComparison option is very useful.

The == and != operators are based on the value, so they are safe to use, even though String is a reference type.

What use does the == operator have for String?

Imagine a thread-safe Queue<String> acting as a communication channel between a producer thread and a consumer thread. It seems perfectly reasonable to use a special String to indicate termination.

// Deliberate use of `new` to make sure JVM does not re-use a cached "EOT".
private static final String EOT = new String("EOT");
...
// Signal we're done.
queue.put(EOT);


// Meanwhile at the consumer end of the queue.
String got = queue.get();
if ( got == EOT ) {
  // Tidy shutdown
}

note that this would be resilient to:

queue.put("EOT");

because "EOT" != EOT even though "EOT".equals(EOT) would be true.

What is the difference between == and equals() in Java?

In general, the answer to your question is "yes", but...

• .equals(...) will only compare what it is written to compare, no more, no less.
• If a class does not override the equals method, then it defaults to the equals(Object o) method of the closest parent class that has overridden this method.
• If no parent classes have provided an override, then it defaults to the method from the ultimate parent class, Object, and so you're left with the Object#equals(Object o) method. Per the Object API this is the same as ==; that is, it returns true if and only if both variables refer to the same object, if their references are one and the same. Thus you will be testing for object equality and not functional equality.
• Always remember to override hashCode if you override equals so as not to "break the contract". As per the API, the result returned from the hashCode() method for two objects must be the same if their equals methods show that they are equivalent. The converse is not necessarily true.

Java, what's the difference between == and equals

Yes, == with objects is a reference comparison* (checks if the operands are references to the same object), while equals is whatever the class involved defines it to mean (within the requirements documented for equals). Some classes define equals as being the same as ==, including Java's arrays. (Because they don't override the default implementation from Object.equals, which uses ==.)

If you want to compare Java arrays based on the equality of their contents, you use Arrays.equals instead.

Your experiment would have worked if you used a class that defined equals in a useful way, you were just unlucky picking arrays. It's a bit tricky to find a class in the JVM to use for this experiment, because so many either don't implement equals (like arrays) or could be confusing because there are several immutable classes which may reuse instances (although not if you explicitly use new; but I don't want to go down a path of having you use new with something you probably shouldn't, like String; more on that here). I'm going to give up on picking a good example and use the slightly old class SimpleDateFormat:

DateFormat a = new SimpleDateFormat("yyyy-MM-dd");
DateFormat b = new SimpleDateFormat("yyyy-MM-dd");
System.out.println(a == b ? "== Same" : "== Different");
System.out.println(a.equals(b) ? "equals Same" : "equals Different");

That outputs

== Different
equals Same

because SimpleDateFormat defines equals to check that the other object is also a SimpleDateFormat with the same formatting.

Live example

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Re your comment on the question:

I have someone the answer points but I only get the == part, if .equals is checking the content, how come the code didnt print "same" for the second if

Because equals doesn't, necessarily, check content. It only does if a class overrides the default Object.equals (which just uses ==) and implements a content check. Arrays don't. (One could argue that they should have, but they don't.) Other classes, like SimpleDateFormat and String and Integer and HashMap do.

In essence: == is always a reference comparison. equals may or may not be a contents comparison, depending on what class you're using it on.

So for instance, say we have this class:

class Example1
{
    private int content;

    Example1(int content) {
        this.content = content;
    }

    public static void main (String[] args) throws java.lang.Exception
    {
        Example1 a = new Example1(42);
        Example1 b = new Example1(42);
        System.out.println(a == b ? "== Same" : "== Different");
        System.out.println(a.equals(b) ? "equals Same" : "equals Different");
    }
}

Since that class doesn't override equals, you'll get the same answer ("Different") for both == and equals. Live Example.

But if we override equals to define what it would mean for two instances to be equal (in this case: because they have the same content):

class Example2
{
    private int content;

    Example2(int content) {
        this.content = content;
    }

    @Override
    public boolean equals(Object obj) {
        if (obj == null || !obj.getClass().equals(this.getClass())) {
            return false;
        }
        Example2 other = (Example2)obj;
        return this.content == other.content;
    }

    @Override
    public int hashCode() {
        return this.content;
    }

    public static void main (String[] args) throws java.lang.Exception
    {
        Example2 a = new Example2(42);
        Example2 b = new Example2(42);
        System.out.println(a == b ? "== Same" : "== Different");
        System.out.println(a.equals(b) ? "equals Same" : "equals Different");
    }
}

Now equals says two instances with the same content are the same, because the class defines what that means. Live Example. (Also note that when overriding equals, you must override hashCode, which is why I've done so above.)

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* More generally, == tests if the values of its operands are the same. The value in the case of reference types is an object reference, and two object reference values are only the same when they refer to the same object are are only different when they refer to different objects (or one of them is null; not referring to an object at all).

C# difference between == and Equals()

When == is used on an expression of type object, it'll resolve to System.Object.ReferenceEquals.

Equals is just a virtual method and behaves as such, so the overridden version will be used (which, for string type compares the contents).

Difference between == operator and Equals() method in C#?

There are several things going on. Firstly, in this example:

string s1 = "a";
string s2 = "a";
Console.WriteLine(s1 == s2);

You claim that:

Both are different object reference.

That's not true due to string interning. s1 and s2 are references to the same object. The C# specification guarantees that - from section 2.4.4.5 of the C# 4 specification:

When two or more string literals that are equivalent according to the string equality operator (§7.10.7) appear in the same program, these string literals refer to the same string instance.

So in this particular case, you would still get "true" even if you printed object.ReferenceEquals(s1, s2), or if you made it use a true reference identity comparison with ==:

object s1 = "a";
object s2 = "a";
Console.WriteLine(s1 == s2); // Still prints True due to string literal interning

However, even if these were references to separate objects, == is overloaded for string. Overloading is a compile-time decision - the implementation to use depends on the compile-time types of the operands. So for example:

string a = new string('x', 1);
string b = new string('x', 1);
Console.WriteLine(a == b); // Uses string's implementation, prints True

object c = a;
object d = b;
Console.WriteLine(c == d); // Reference identity comparison, prints False

Compare that with object.Equals(object) which is a virtual method. As it happens, String overloads this method as well, but importantly it overrides it. So if we change our code to:

string a = new string('x', 1);
string b = new string('x', 1);
Console.WriteLine(a.Equals((object) b));

object c = a;
object d = b;
Console.WriteLine(c.Equals(d));

... then both method calls in the compiled code will simply be to object.Equals(object), but they'll still both print True because of polymorphism: the implementation in String will be used.

Here's what a call to the overloaded method would look like:

string a = new string('x', 1);
string b = new string('x', 1);
Console.WriteLine(a.Equals(b)); // Calls string.Equals(string)

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