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.....
This question already has answers here:
How do I compare strings in Java?
(23 answers)
What is the difference between "text" and new String("text")?
(13 answers)
Closed 3 years ago.
public class Test {
public static void main(String[] args)
{
String s1 = "HELLO";
String s2 = "HELLO";
System.out.println(s1 == s2); // true
}
}
But when I use :
public class Test {
public static void main(String[] args)
{
String s1 = new String("HELLO");
String s2 = new String("HELLO");
System.out.println(s1 == s2); // false
}
}
Can anybody please explain the difference here? Thankyou!
In the first example
String s1 = "HELLO";
String s2 = "HELLO";
the values of s1 and s2 are compile-time constants. Thus, the compiler does only generate a single String-object, holding the value "HELLO" and assings it to both s1 and s2. This is a special case of Common Subexpression Elimination, a well-known compiler optimization. Thus s1 == s2 returns true.
In the second example, two different Strings are constructed explicitly through new. Thus, they have to be separate objects per the semantics of new.
I created an Ideone demo a while back that highlights some cases that show this behaviour.
You can enforce that the same String is return by using String::intern():
String s1 = new String("HELLO").intern();
String s2 = new String("HELLO").intern();
System.out.println(s1 == s2); // will print "true";
Ideone demo
In case of String literal,before creating new Object in String Constant Pool ,JVM will check already same Object persist in SCP area or not if yes it will point to same object instead of creating new Object.Hence, below code s1 == s2 is true
String s1 = "HELLO";
String s2 = "HELLO";
System.out.println(s1 == s2); // true
but we are creating new object by using new keyword, it will create object in heap area, hence s1 and s2 are pointing to two different object, hence it is return false
== tests for reference equality (whether they are the same object).
.equals() tests for value equality (whether they are logically "equal").
from here How do I compare strings in Java?
== compares the objects reference pointer. When 2 objects are same exact object it will be true.
Instantiating a string using double quotes uses the string pool, creates a string once and reuses it.
Instantiating a string wit new always creates a brand new string.
== tests for reference equality (whether they are the same object).
First Case
System.out.println(s1 == s2); // true
Because you are comparing literals that are interned by the compiler and thus refer to the same object. Moreover, a string literal always refers to the same instance of class String. This is because 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.
Second Case
System.out.println(s1 == s2); // false
You are comparing the Object reference which is different so you are getting false.
Please check https://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.28
i first case u comparing two strings with its ASCII values. thats why...//true
and in second case you are comparing two functions/methods. thats why... //false
the first one is true because s1 and s2 refer to the same string literal in the method area, the memory references are the same. ( == checks just the references in string). when the same string literal is created more than once, only one copy of each distinct string value is stored.
the second one is false because s1 and s2 refer to two different objects in the heap. different objects always have different memory references.
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.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How do I compare strings in Java?
I am new to Java and I have difficulties in understanding String comparison. Can anyone explain the differences between the following scenarios?
Scenario 1 :
String a = "abc";
String b = "abc";
When I run the if(a == b) it returns true.
Scenario 2 :
String a = new String("abc");
String b = new String("abc");
and run if(a == b) then it returns false.
What is the difference?
== operator compares the references of two objects in memory. If they point to the same location then it returns true.String object in java are Immutable, so when you create Strings like in scenario1 then it didn't create new string. It just points the second string to the memory location of first string.
However, .equals() method compares the content of the String. When strings has same value then this method returns true.
So, in general it is recommended to use equals() method instead of ==.
It's because of Java String constant memory pool. Same valued literals are stored once.
String a = "abc";
String b = "abc";
// Now there is 1 string ("abc") and 2 references pointing to it.
String a = new String("abc");
String b = new String("abc");
// Now you have 2 string instances and 2 references.
Scenario 1 returns true because of a compiler optimization.
In general you should use equals() instead of == to compare strings.
In Java you need to use like this if(str.equals(str2)) which compares actual value of string rather than references.
Case1:
String a = "abc";
String b = "abc";
if(a == b)
In this case abc is cached in String constant pool thus a new string is not created String b = "abc"; b just refers to the string created by a it returns true as a and b both point to the same Object in the memory.
Case2:
String a = new String("abc");
String b = new String("abc");
and run if(a == b) then it returns false.
Here a two Strings are created and == operator just checks if two references point to the same reference, which it doesnt in this case thus it returns false
Two ways of creating string are
1) String s ="hello"; No. of string literal = 1 i.e. "hello" - No. of String objects on heap = 0
2) String s= new String("hello"); - No. of string literals =1 and No. of string objects =1
Java maintains a string pool of "LITERALS" and objects are going to stay on heap.
Advantage of string pooling: 1) Reduced memory usage*(PermGenSpace Issue) 2) Faster Comparision i.e == comparision 3) Faster lookup
Disadvantages: 1) Overhead of maintaining pool
How to pool String objects? Use Intern() on a string to add it to the pool. Downside of interning: 1) You may forget to intern some strings and compare them by == leading to unexpected results.
The reason is that the String literal "abc" will be turned into a global String instance for all its ocurrences, it will be the same String instance therefore you can be sure that "abc" == "abc". It is possible for the compiler to do that because String instances are immutable. However, if you explicitly allocate the String they will be two different instances and they will also be different to the String instance implicitly created by the compiler i.e. new String("abc") != new String("abc") and "abc" != new String("abc").
Another good example to understand what the compiler is doing is to look at this code:
"abc".contains("a");
you see that the literal behaves like an instance of a String type. You may exploit this to minimize programming errors e.g.
// this is OK and the condition will evaluate to false
String myStringValue = null;
if ("abc".equals(myStringValue)) { // false
whereas this code results in NPE:
// this will produce a NPE
String myStringValue = null;
if (myStringValue.equals("abc")) { // NPE
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Java string comparison?
I was trying to do this:
boolean exit = false;
while(exit==false && convStoreIndex<convStoreLength) {
if(conversionStore[convStoreIndex].getInUnit()==inUnit) {
indexCount++;
exit=true;
}
convStoreIndex++;
}
but the if-condition never went true, even if the two Strings were the same(checked this in the debugger).
so I added some lines:
boolean exit = false;
while(exit==false && convStoreIndex<convStoreLength) {
Log.v("conversionStore["+String.valueOf(convStoreIndex)+"]", conversionStore[convStoreIndex].getInUnit()+"|"+inUnit);
String cs = conversionStore[convStoreIndex].getInUnit();
String iu = inUnit;
Log.v("cs", cs);
Log.v("iu", iu);
Log.v("Ergebnis(cs==iu)", String.valueOf(cs==iu));
if(conversionStore[convStoreIndex].getInUnit()==inUnit) {
indexCount++;
exit=true;
}
convStoreIndex++;
}
and here is the extract from LogCat:
09-15 11:07:14.525: VERBOSE/cs(585): kg
09-15 11:07:16.148: VERBOSE/iu(585): kg
09-15 11:07:17.687: VERBOSE/Ergebnis(cs==iu)(585): false
the class of conversionStore:
class ConversionStore {
private String inUnit;
[...]
public String getInUnit() {
return inUnit;
}
}
Who is going crazy, java or me?
Don't use == to compare String objects, use .equals():
if(conversionStore[convStoreIndex].getInUnit().equals(inUnit)) {
To compare Strings for equality, don't use ==. The == operator checks
to see if two objects are exactly the same object. Two strings may be
different objects, but have the same value (have exactly the same
characters in them). Use the .equals() method to compare strings for
equality.
Straight from the first link Google provided when searching "Java string comparison"...
Please use String.equals() to compare by string content instead of reference identity.
You are comparing your Strings using == : cs==iu.
But this will return true only if both Strings are actually the same object. This is not the case here: you have two distinct instances of String that contain the same value.
You should use String.compareTo(String).
use .equals(); method instread of ==
Becase ::
They both differ very much in their significance. equals() method is present in the java.lang.Object class and it is expected to check for the equivalence of the state of objects! That means, the contents of the objects. Whereas the '==' operator is expected to check the actual object instances are same or not.
For example, lets say, you have two String objects and they are being pointed by two different reference variables s1 and s2.
s1 = new String("abc");
s2 = new String("abc");
Now, if you use the "equals()" method to check for their equivalence as
if(s1.equals(s2))
System.out.println("s1.equals(s2) is TRUE");
else
System.out.println("s1.equals(s2) is FALSE");
You will get the output as TRUE as the 'equals()' method check for the content equivality.
Lets check the '==' operator..
if(s1==s2)
System.out.printlln("s1==s2 is TRUE");
else
System.out.println("s1==s2 is FALSE");
Now you will get the FALSE as output because both s1 and s2 are pointing to two different objects even though both of them share the same string content. It is because of 'new String()' everytime a new object is created.
Try running the program without 'new String' and just with
String s1 = "abc";
String s2 = "abc";
You will get TRUE for both the tests.
Reson::
After the execution of String str1 = “Hello World!”; the JVM adds the string “Hello World!” to the string pool and on next line of the code, it encounters String str2 =”Hello World!”; in this case the JVM already knows that this string is already there in pool, so it does not create a new string. So both str1 and str2 points to the same string means they have same references. However, str3 is created at runtime so it refers to different string.
Two things:
You don't need to write "boolean==false" you can just write "!boolean" so in your example that would be:
while(!exit && convStoreIndex<convStoreLength) {
Second thing:
to compare two Strings use the String.equals(String x) method, so that would be:
if(conversionStore[convStoreIndex].getInUnit().equals(inUnit)) {