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Substring in if statement [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
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");
}
}

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.

Meet Jorman
Jorman is a successful businessman and has 2 houses.
But others don't know that.
Is it the same Jorman?
When you ask neighbours from either Madison or Burke streets, this is the only thing they can say:
Using the residence alone, it's tough to confirm that it's the same Jorman. Since they're 2 different addresses, it's just natural to assume that those are 2 different persons.
That's how the operator == behaves. So it will say that datos[0]==usuario is false, because it only compares the addresses.
An Investigator to the Rescue
What if we sent an investigator? We know that it's the same Jorman, but we need to prove it. Our detective will look closely at all physical aspects. With thorough inquiry, the agent will be able to conclude whether it's the same person or not. Let's see it happen in Java terms.
Here's the source code of String's equals() method:
It compares the Strings character by character, in order to come to a conclusion that they are indeed equal.
That's how the String equals method behaves. So datos[0].equals(usuario) will return true, because it performs a logical comparison.

It's good to notice that in some cases use of "==" operator can lead to the expected result, because the way how java handles strings - string literals are interned (see String.intern()) during compilation - so when you write for example "hello world" in two classes and compare those strings with "==" you could get result: true, which is expected according to specification; when you compare same strings (if they have same value) when the first one is string literal (ie. defined through "i am string literal") and second is constructed during runtime ie. with "new" keyword like new String("i am string literal"), the == (equality) operator returns false, because both of them are different instances of the 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)
Update: 01.04.2013 I updated this post due comments below which are somehow right. Originally I declared that interning (String.intern) is side effect of JVM optimization. Although it certainly save memory resources (which was what i meant by "optimization") it is mainly feature of language

The == operator checks if the two references point to the same object or not.
.equals() checks for the actual string content (value).
Note that the .equals() method belongs to class Object (super class of all classes). You need to override it as per you class requirement, but for String it is already implemented and it checks whether two strings have the same value or not.
Case1)
String s1 = "Stack Overflow";
String s2 = "Stack Overflow";
s1 == s1; // true
s1.equals(s2); // true
Reason: String literals created without null are stored in the string pool in the permgen area of the heap. So both s1 and s2 point to the same object in the pool.
Case2)
String s1 = new String("Stack Overflow");
String s2 = new String("Stack Overflow");
s1 == s2; // false
s1.equals(s2); // true
Reason: If you create a String object using the `new` keyword a separate space is allocated to it on the heap.

equals() function is a method of Object class which should be overridden by programmer. String class overrides it to check if two strings are equal i.e. in content and not reference.
== operator checks if the references of both the objects are the same.
Consider the programs
String abc = "Awesome" ;
String xyz = abc;
if(abc == xyz)
System.out.println("Refers to same string");
Here the abc and xyz, both refer to same String "Awesome". Hence the expression (abc == xyz) is true.
String abc = "Hello World";
String xyz = "Hello World";
if(abc == xyz)
System.out.println("Refers to same string");
else
System.out.println("Refers to different strings");
if(abc.equals(xyz))
System.out.prinln("Contents of both strings are same");
else
System.out.prinln("Contents of strings are different");
Here abc and xyz are two different strings with the same content "Hello World". Hence here the expression (abc == xyz) is false where as (abc.equals(xyz)) is true.
Hope you understood the difference between == and <Object>.equals()
Thanks.

== tests for reference equality.
.equals() tests for value equality.
Consequently, if you actually want to test whether two strings have the same value you should use .equals() (except in a few situations where you can guarantee that two strings with the same value will be represented by the same object eg: String interning).
== is for testing whether two strings are the same Object.
// These two have the same value
new String("test").equals("test") ==> true
// ... but they are not the same object
new String("test") == "test" ==> false
// ... neither are these
new String("test") == new String("test") ==> false
// ... but these are because literals are interned by
// the compiler and thus refer to the same object
"test" == "test" ==> true
// concatenation of string literals happens at compile time resulting in same objects
"test" == "te" + "st" ==> true
// but .substring() is invoked at runtime, generating distinct objects
"test" == "!test".substring(1) ==> false
It is important to note that == is much cheaper than equals() (a single pointer comparision instead of a loop), thus, in situations where it is applicable (i.e. you can guarantee that you are only dealing with interned strings) it can present an important performance improvement. However, these situations are rare.

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.

Let's analyze the following Java, to understand the identity and equality of Strings:
public static void testEquality(){
String str1 = "Hello world.";
String str2 = "Hello world.";
if (str1 == str2)
System.out.print("str1 == str2\n");
else
System.out.print("str1 != str2\n");
if(str1.equals(str2))
System.out.print("str1 equals to str2\n");
else
System.out.print("str1 doesn't equal to str2\n");
String str3 = new String("Hello world.");
String str4 = new String("Hello world.");
if (str3 == str4)
System.out.print("str3 == str4\n");
else
System.out.print("str3 != str4\n");
if(str3.equals(str4))
System.out.print("str3 equals to str4\n");
else
System.out.print("str3 doesn't equal to str4\n");
}
When the first line of code String str1 = "Hello world." executes, a string \Hello world."
is created, and the variable str1 refers to it. Another string "Hello world." will not be created again when the next line of code executes because of optimization. The variable str2 also refers to the existing ""Hello world.".
The operator == checks identity of two objects (whether two variables refer to same object). Since str1 and str2 refer to same string in memory, they are identical to each other. The method equals checks equality of two objects (whether two objects have same content). Of course, the content of str1 and str2 are same.
When code String str3 = new String("Hello world.") executes, a new instance of string with content "Hello world." is created, and it is referred to by the variable str3. And then another instance of string with content "Hello world." is created again, and referred to by
str4. Since str3 and str4 refer to two different instances, they are not identical, but their
content are same.
Therefore, the output contains four lines:
Str1 == str2
Str1 equals str2
Str3! = str4
Str3 equals str4

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

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 they 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");
}

Generally .equals is used for Object comparison, where you want to verify if two Objects have an identical value.
== for reference comparison (are the two Objects the same Object on the heap) & to check if the Object is null. It is also used to compare the values of primitive types.

== operator compares the reference of an object in Java. You can use string's equals method .
String s = "Test";
if(s.equals("Test"))
{
System.out.println("Equal");
}

If you are going to compare any assigned value of the string i.e. primitive string, both "==" and .equals will work, but for the new string object you should use only .equals, and here "==" will not work.
Example:
String a = "name";
String b = "name";
if(a == b) and (a.equals(b)) will return true.
But
String a = new String("a");
In this case if(a == b) will return false
So it's better to use the .equals operator...

The == operator is a simple comparison of values.
For object references the (values) are the (references). So x == y returns true if x and y reference the same object.

I know this is an old question but here's how I look at it (I find very useful):
Technical explanations
In Java, all variables are either primitive types or references.
(If you need to know what a reference is: "Object variables" are just pointers to objects. So with Object something = ..., something is really an address in memory (a number).)
== compares the exact values. So it compares if the primitive values are the same, or if the references (addresses) are the same. That's why == often doesn't work on Strings; Strings are objects, and doing == on two string variables just compares if the address is same in memory, as others have pointed out. .equals() calls the comparison method of objects, which will compare the actual objects pointed by the references. In the case of Strings, it compares each character to see if they're equal.
The interesting part:
So why does == sometimes return true for Strings? Note that Strings are immutable. In your code, if you do
String foo = "hi";
String bar = "hi";
Since strings are immutable (when you call .trim() or something, it produces a new string, not modifying the original object pointed to in memory), you don't really need two different String("hi") objects. If the compiler is smart, the bytecode will read to only generate one String("hi") object. So if you do
if (foo == bar) ...
right after, they're pointing to the same object, and will return true. But you rarely intend this. Instead, you're asking for user input, which is creating new strings at different parts of memory, etc. etc.
Note: If you do something like baz = new String(bar) the compiler may still figure out they're the same thing. But the main point is when the compiler sees literal strings, it can easily optimize same strings.
I don't know how it works in runtime, but I assume the JVM doesn't keep a list of "live strings" and check if a same string exists. (eg if you read a line of input twice, and the user enters the same input twice, it won't check if the second input string is the same as the first, and point them to the same memory). It'd save a bit of heap memory, but it's so negligible the overhead isn't worth it. Again, the point is it's easy for the compiler to optimize literal strings.
There you have it... a gritty explanation for == vs. .equals() and why it seems random.

#Melkhiah66 You can use equals method instead of '==' method to check the equality.
If you use intern() then it checks whether the object is in pool if present then returns
equal else unequal. equals method internally uses hashcode and gets you the required result.
public class Demo
{
public static void main(String[] args)
{
String str1 = "Jorman 14988611";
String str2 = new StringBuffer("Jorman").append(" 14988611").toString();
String str3 = str2.intern();
System.out.println("str1 == str2 " + (str1 == str2)); //gives false
System.out.println("str1 == str3 " + (str1 == str3)); //gives true
System.out.println("str1 equals str2 " + (str1.equals(str2))); //gives true
System.out.println("str1 equals str3 " + (str1.equals(str3))); //gives true
}
}

The .equals() will check if the two strings have the same value and return the boolean value where as the == operator checks to see if the two strings are the same object.

Someone said on a post higher up that == is used for int and for checking nulls.
It may also be used to check for Boolean operations and char types.
Be very careful though and double check that you are using a char and not a String.
for example
String strType = "a";
char charType = 'a';
for strings you would then check
This would be correct
if(strType.equals("a")
do something
but
if(charType.equals('a')
do something else
would be incorrect, you would need to do the following
if(charType == 'a')
do something else

a==b
Compares references, not values. The use of == with object references is generally limited to the following:
Comparing to see if a reference is null.
Comparing two enum values. This works because there is only one object for each enum constant.
You want to know if two references are to the same object
"a".equals("b")
Compares values for equality. Because this method is defined in the Object class, from which all other classes are derived, it's automatically defined for every class. However, it doesn't perform an intelligent comparison for most classes unless the class overrides it. It has been defined in a meaningful way for most Java core classes. If it's not defined for a (user) class, it behaves the same as ==.

Use Split rather than tokenizer,it will surely provide u exact output
for E.g:
string name="Harry";
string salary="25000";
string namsal="Harry 25000";
string[] s=namsal.split(" ");
for(int i=0;i<s.length;i++)
{
System.out.println(s[i]);
}
if(s[0].equals("Harry"))
{
System.out.println("Task Complete");
}
After this I am sure you will get better results.....

String pool memory alloaction

String s = "abc";
String s4 = s + "";
System.out.println(s4 == s);
System.out.println(s4.equals(s));
This prints:
false
true
Can anybody please explain why is it so?

String s="abc"; // goes to string constants pool
String s4 = s +"" ;// has value "abc" but goes on heap and not in string constants pool
System.out.println(s4==s);//well,, the references are not equal. There are 2 diffrent instances with value="abc"
System.out.println(s4.equals(s)); // equals() checks for value and the values of both insatnces are equal.

Here is your code with output in comment -
String s="abc";
String s4 = s +"" ;
System.out.println(s4==s); //false
System.out.println(s4.equals(s)); //true
First one is false because it is checking the reference of s4 and s. Since these two reference are different it evaluated to false. Here == (the equality) operator, when applied on reference type, is used to check whether two reference are same or different.
equals() is a method which is achieved by every reference type from object class. You can override the equals() method for your own class.
The equals(s) method is used to check whether the two object are meaningfully equals or not. For String class the meaningfully equals means the two comparing strings are same but their references may be different. String class has already override the equals()method and hence you need not override the equals() method by yourself.
Now for understanding string pool concept, consider the following code snippet -
String s5 = s;
System.out.println(s5); //abc
System.out.println(s5==s); //true
System.out.println(s5.equals(s)); //true
Here s5==s evaluated to true. Because s5 is not referencing a new String object. It's referencing the the existing String object (that is - "abc" referenced by s) in the string pool. Now the both reference s5 and s4 are equals. In the last statement s5 and s are meaningfully equals and hence s5.equals(s) evaluated to true.
Hope it will help.
Thanks a lot.

Java String Objects [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
I have the following code:
public class Porow{
public static void main (String[] args){
String s1 = "foo";
String s2 = "foo";
System.out.println(s1 == s2);
System.out.println(s1.equals(s2));
String s3 = new String ("foo");
System.out.println(s1 == s3);
System.out.println(s1.equals(s3));
}
}
which outputs:
true
true
false
true
I am trying to understand and explain what's going on here. And I need some guidance. Will I be right to say s1 and s2 are same objects stored in different parts of memory and that's why we get the first 2 trues?
Why is the 3rd output false? is s3 a different object? and why does it produce true with the equals method on s1? I'd appreciate some help. Thanks :).
EDIT:
This question is not about == nor equals() I know what they are. My question is more related to memory references and addresses. So don't just presume I'm asking about String Comparison. I'm asking about a whole different thing here.

That's because in the 3rd scenario, a new object is created(the new keyword creates a new string object) and == is used for object reference equality. That's why it returns false.
Whereas the equals() method is used for String value equality, and hence, it returns true.
In the first and second case, the "foo" String literal is interned and thus, both s1 and s2 refer the same String literal.

The == operator checks reference equivalence. If the two references are equivalent, then they're the same instance.
Your first example works because the strings are interned - but don't rely on this. equals() actually checks to see if the two Strings contain the same data.
With the new keyword, a new reference is created for the other string - hence, they no longer point to the same instance.

Because == tests that the strings refer to the same object and equals tests if they are equivalent. Your line
String s3 = new String ("foo");
constructs a new reference to the constant "foo" String. Take a look at What is String interning? for a bit more, but be aware that interning will use PermGen space.

String are put into a String pool
When you have
String s1 = "foo";
String s2 = "foo";
Only one "foo" is placed into the String pool.
So
s1 == s2 is true becuase they reference the Same String "foo" in the pool
To answer your question, your assumptions are right!
The last true s1.equals(s3) compares the value "foo", that's why they are equal.
So just remember with Strings
== compares Object reference
equals compares value.

== operator with Strings

The code below should not print "Bye", since == operator is used to compare references, but oddly enough, "Bye" is still printed. Why does this happen? I'm using Netbeans 6.9.1 as the IDE.
public class Test {
public static void main(String [] args) {
String test ="Hi";
if(test=="Hi"){
System.out.println("Bye");
}
}
}

This behavior is because of interning. The behavior is described in the docs for String#intern (including why it's showing up in your code even though you never call String#intern):
A pool of strings, initially empty, is maintained privately by the class String.
When the intern method is invoked, if the pool already contains a string equal to this String object as determined by the equals(Object) method, then the string from the pool is returned. Otherwise, this String object is added to the pool and a reference to this String object is returned.
It follows that for any two strings s and t, s.intern() == t.intern() is true if and only if s.equals(t) is true.
All literal strings and string-valued constant expressions are interned. String literals are defined in §3.10.5 of the Java Language Specification.
So for example:
public class Test {
private String s1 = "Hi";
public static void main(String [] args) {
new Test().test();
System.exit(0);
}
public void test() {
String s2 ="Hi";
String s3;
System.out.println("[statics] s2 == s1? " + (s2 == s1));
s3 = "H" + part2();
System.out.println("[before interning] s3 == s1? " + (s3 == s1));
s3 = s3.intern();
System.out.println("[after interning] s3 == s1? " + (s3 == s1));
System.exit(0);
}
protected String part2() {
return "i";
}
}
Output:
[statics] s2 == s1? true
[before interning] s3 == s1? false
[after interning] s3 == s1? true
Walking through that:
The literal assigned to s1 is automatically interned, so s1 ends up referring to a string in the pool.
The literal assigned to s2 is also auto-interned, and so s2 ends up pointing to the same instance s1 points to. This is fine even though the two bits of code may be completely unknown to each other, because Java's String instances are immutable. You can't change them. You can use methods like toLowerCase to get back a new string with changes, but the original you called toLowerCase (etc.) on remains unchanged. So they can safely be shared amongst unrelated code.
We create a new String instance via a runtime operation. Even though the new instance has the same sequence of characters as the interned one, it's a separate instance. The runtime doesn't intern dynamically-created strings automatically, because there's a cost involved: The work of finding the string in the pool. (Whereas when compiling, the compiler can take that cost onto itself.) So now we have two instances, the one s1 and s2 point to, and the one s3 points to. So the code shows that s3 != s1.
Then we explicitly intern s3. Perhaps it's a large string we're planning to hold onto for a long time, and we think it's likely that it's going to be duplicated in other places. So we accept the work of interning it in return for the potential memory savings. Since interning by definition means we may get back a new reference, we assign the result back to s3.
And we can see that indeed, s3 now points to the same instance s1 and s2 point to.

Hard-coded Strings are compiled into the JVM's String Table, which holds unique Strings - that is the compiler stores only one copy of "Hi", so you are comparing that same object, so == works.
If you actually create a new String using the constructor, like new String("Hi"), you will get a different object.

There is a String cache in java. Where as in this case the same object is returned from the cache which hold the same reference.

The main reason for that is because the "Hi" is picked up from String Pool. The immutable object must have some sort of cache so it can perform better. So String class is immutable and it uses String Pool for basic cache.
In this case, "Hi" is in the String pool and all the String having values of "Hi" will have the same reference for String Pool.

Java == for String objects ceased to work?

public class Comparison {
public static void main(String[] args) {
String s = "prova";
String s2 = "prova";
System.out.println(s == s2);
System.out.println(s.equals(s2));
}
}
outputs:
true
true
on my machine. Why? Shouldn't be == compare object references equality?

Because String instances are immutable, the Java language is able to make some optimizations whereby String literals (or more generally, String whose values are compile time constants) are interned and actually refer to the same (i.e. ==) object.
JLS 3.10.5 String Literals
Each string literal is a reference to an instance of class String. String objects have a constant value. String literals-or, more generally, strings that are the values of constant expressions -are "interned" so as to share unique instances, using the method String.intern.
This is why you get the following:
System.out.println("yes" == "yes"); // true
System.out.println(99 + "bottles" == "99bottles"); // true
System.out.println("7" + "11" == "" + '7' + '1' + (char) (50-1)); // true
System.out.println("trueLove" == (true + "Love")); // true
System.out.println("MGD64" == "MGD" + Long.SIZE);
That said it needs to be said that you should NOT rely on == for String comparison in general, and should use equals for non-null instanceof String. In particular, do not be tempted to intern() all your String just so you can use == without knowing how string interning works.
Related questions
Java String.equals versus ==
difference between string object and string literal
what is the advantage of string object as compared to string literal
Is it good practice to use java.lang.String.intern()?
On new String(...)
If for some peculiar reason you need to create two String objects (which are thus not == by definition), and yet be equals, then you can, among other things, use this constructor:
public String(String original) : Initializes a newly created String object so that it represents the same sequence of characters as the argument; in other words, the newly created string is a copy of the argument string. Unless an explicit copy of original is needed, use of this constructor is unnecessary since Strings are immutable.
Thus, you can have:
System.out.println("x" == new String("x")); // false
The new operator always create a new object, thus the above is guaranteed to print false. That said, this is not generally something that you actually need to do. Whenever possible, you should just use string literals instead of explicitly creating a new String for it.
Related questions
Java Strings: “String s = new String(”silly“);”
What is the purpose of the expression “new String(…)” in Java?

JLS, 3.10.5 => It is guaranteed that a literal string object will be reused by any other code running in the same virtual machine that happens to contain the same string literal

If you explicitly create new objects, == returns false:
String s1 = new String("prova");
String s2 = new String("prova");
System.out.println(s1 == s2); // returns false.
Otherwise the JVM can use the same object, hence s1 == s2 will return true.

It does. But String literals are pooled, so "prova" returns the same instance.

String s = "prova";
String s2 = "prova";
s and s2 are literal strings which are pointing the same object in String Pool of JVM, so that the comparison returns true.

Yes, "prova" is stored in the java inner string pool, so its the same reference.

Source code literals are part of a constant pool, so if the same literal appears multiple times, it will be the same object at runtime.

The JVM may optimize the String usage so that there is only one instance of the "equal" String in memory. In this case also the == operator will return true. But don't count on it, though.

You must understand that "==" compares references and "equals" compares values. Both s and s1 are pointing to the same string literal, so their references are the same.

When you put a literal string in java code, the string is automatically interned by the compiler, that is one static global instance of it is created. Or more specifically, it is put into a table of interned strings. Any other quoted string that is exactly the same, content-wise, will reference the same interned string.
So in your code s and s2 are the same string

Ideally it should not happen ever. Because java specification guarantees this. So I think it may be the bug in JVM, you should report to the sun microsystems.