We have Integer class in JAVA, but I couldn't find any equivalent class in C#? Does c# have any equivalent? If not, how do I get JAVA Integer class behavior in c#?
Why do I need this?
It is because I'm trying to migrate JAVA code to c# code. If there is an equivalent way, then code migration would be easier. To addon, I need to store references of the Integer and I don't think I can create reference of int or Int32.
C# has a unified type system, so int can be implicitly boxed into an object reference. The only reason Integer exists in Java is so that it can be converted to an object reference and stored in references to be used in other container classes.
Since C# can do that without another type, there's no corresponding class to Integer.
Code migration won´t work out of the box for any type of language without any manual changes. There are things such as a class Integer that simply does not exist within (C# why should it anyway, see recursives answer), so you´d have to do some work on your own. The nearest equivalent to what you´re after is Int32 or its alias int. However you may of course write your own wrapper-class:
public class Integer
{
public int Value { get; set; }
public Integer() { }
public Integer( int value ) { Value = value; }
// Custom cast from "int":
public static implicit operator Integer( Int32 x ) { return new Integer( x ); }
// Custom cast to "int":
public static implicit operator Int32( Integer x ) { return x.Value; }
public override string ToString()
{
return string.Format( "Integer({0})", Value );
}
}
The beauty of C# is that it has a unified type system. Everything derives from object, even primitive types. Because of this, all keywords are simply aliases for a corresponding class or struct. Java does not use a unified type system, so a separate Integer class is required to wrap the int primitive. In C# int is synonym for the Int32 struct.
What you're looking for has been right in front of you the whole time. Start using the dot notation directly on the int keyword (i.e. int.whatever()) to access the all goodness of the .NET version of the Javian Integer class.
I did some testing with Nullable types in a console application and it appears that they do not behave as you wish. For example:
static void Main(string[] args)
{
int? x = 1;
Foo(ref x);
Console.WriteLine(x);//Writes 2
}
private static void Foo(ref int? y)
{
y += 1;
var l = new List<int?>();
l.Add(y);
l[0] += 1;//This does not affect the value of x devlared in Main
Console.WriteLine(l[0]);//Writes 3
Console.WriteLine(y);//writes 2
Foo2(l);
}
private static void Foo2(List<int?> l)
{
l[0] += 1;
Console.WriteLine(l[0]);//writes 4
}
But if you roll your own generic class to wrap primitive/value types for use within your application you can get the behavior you are expecting:
public class MyType<T>
{
public T Value { get; set; }
public MyType() : this(default(T))
{}
public MyType(T val)
{
Value = val;
}
public override string ToString()
{
return this.Value.ToString();
}
}
static void Main(string[] args)
{
var x = new MyType<int>(1);
Foo(x);
Console.WriteLine(x);//Writes 4
}
private static void Foo(MyType<int> y)
{
y.Value += 1;
var l = new List<MyType<int>>();
l.Add(y);
l[0].Value += 1;//This does affect the value of x devlared in Main
Console.WriteLine(l[0]);//Writes 3
Console.WriteLine(y);//writes 3
Foo2(l);
}
private static void Foo2(List<MyType<int>> l)
{
l[0].Value += 1;
Console.WriteLine(l[0]);//writes 4
}
int, int? and System.Int32 are all struct and thus value types and does not compare to Java's Integer wrapper class which is a reference type.
System.Object class though a reference type can cause issue as boxing creates immutable object. In short, you can't alter a boxed value.
int a = 20;
Object objA = a; //Boxes a value type into a reference type, objA now points to boxed [20]
Object objB = objA; //Both objA & objB points to boxed [20]
objA = 40; //While objB points to boxed [20], objA points to a new boxed [40]
//Thus, it creates another ref type boxing a 40 value integer value type,
//Boxed values are immutable like string and above code does not alter value of previous boxed value [20]
Console.WriteLine($"objA = {objA}, objB={objB}");
//Output: objA = 40, objB=20
What exactly corresponds to Java's Integer is a custom generic wrapper class.
int a = 20;
Wrapper<int> wrapA = new Wrapper<int>(a);
Wrapper<int> wrapB = wrapA; //both wrapA and wrapB are pointing to [20]
wrapA.Value = 40; //Changing actual value which both wrapA and wrapB are pointing to
Console.WriteLine($"wrapA = {wrapA}, wrapB={wrapB}");
//Output: wrapA = 40, wrapB=40
Console.ReadKey();
Implementation of the wrapper class is given below:
public class Wrapper<T> where T : struct
{
public static implicit operator T(Wrapper<T> w)
{
return w.Value;
}
public Wrapper(T t)
{
_t = t;
}
public T Value
{
get
{
return _t;
}
set
{
_t = value;
}
}
public override string ToString()
{
return _t.ToString();
}
private T _t;
}
As pointed out in other answers, C# has a unified type system so everything derives from object. If you need to handle null values then use int? to specify that the integer object can be null.
c# have a integer type called int link is here
https://msdn.microsoft.com/en-us/library/5kzh1b5w.aspx
Related
Given this code snippet:
class Ex1{
public static void main(String args[]){
int x = 10;
int y = new Ex1().change(x);
System.out.println(x+y);
}
int change(int x){
x=12;
return x;
}
}
I understand that the x in main won't get changed by the change method and return the value 22 because Java primitives are call-by-value. However, if I change all the int to Integer, making them objects and therefore theoretically call-by-value-of-reference, why does the program still return 22?
Is it possible to modify the method change such that it also modifies the variable x in main?
EDIT: new snippet
class Ex1{
public static void main(String args[]){
Integer x = 10;
Integer y = new Ex1().change(x);
System.out.println(x+y);
}
Integer change(Integer x){
x=12;
return x;
}
}
Both value and reference types are passed by-value in Java (see the Java Tutorials). This means that the passed-in reference still points at the same object as before the call, even if the internals of a method change the reference assigned to a method's parameter variable.
The primitive wrappers are all reference types, so there is no difference between their behaviour and the behaviour of any other reference type when passed as an argument to a method.
However, you can change the values inside a reference object, and those changes will be reflected after the method call completes, in the calling method. You can't do this with the primitive wrappers though: they are immutable.
public static void main(String[] args){
Foo parentFoo = new Foo(1);
System.out.println(parentFoo); // prints "instance 1, data is now 1"
changeReferenceFail(parentFoo); // prints "instance 2, data is now 2"
System.out.println(parentFoo); // prints "instance 1, data is now 1"
mutateReference(parentFoo); // prints "instance 1, data is now 3"
System.out.println(parentFoo); // prints "instance 1, data is now 3"
}
private static void changeReferenceFail(Foo myFoo) {
myFoo = new Foo(2); // assigns a new object to the myFoo parameter variable
System.out.println(myFoo);
}
private static void mutateReference(Foo myFoo) {
myFoo.setData(3); // changes the reference variable internals
System.out.println(myFoo);
}
...
class Foo {
private static int iidSeed = 0;
private int iid = 0;
private int data;
public Foo(int data) {
this.data = data;
this.iid = ++iidSeed;
}
public void setData(int data) { this.data = data; }
public String toString() {
return String.format("instance %d, data is now %d", this.iid, this.data);
}
}
You asked: "Is it possible to modify the method change such that it also modifies the variable x in main?".
You can either pass a reference object, and modify an internal field (as per mutateReference above). Or you can return a new integer and assign it to your local variableexactly as you are doing already.
Integer change(Integer x){
x=12;
return x;
}
Because this does not change what is stored inside the object Integer x, but assigns a new value to the variable x. It is not the original argument object being changed, but a new Integer object is created assigned to the variable formerly holding the original object.
As you said, when passing an object to a function, you actually pass the value of its reference. Thus, statements like myParam = something have no effect on the object passed to the method, only method calls such as myParam.mutate() can change its state. Nevertheless, Integer is an immutable class so you will not be able by any mean, to change the value of the Integer in the main.
You are passing the value of x to your method, that applies that value to another variable x. You would need to modify the correct instance of x to change it in main. This snippet changes x, although I'm sure you knew how to do this already.
class Ex1{
int x = 10;
public static void main(String args[]){
System.out.println(x);
changeX(15);
System.out.println(x);
}
void changeX(int newVal){
x=newVal;
}
}
I know here is no pointer in Java. But how do I change a value in the calling scope? For instance, I want to write a function that takes an integer num, set the integer to 0 if it's greater than 21, otherwise do nothing.
In the main, my code is as follow:
int a=34;
KillOver21(a);
System.out.print(a);
I expect an 0.
Java is pass by value, so a copy of the parameter a is sent to the method, so modification to a in the method will not affect the original argument a in main
The max you can do is return int from KillOver21(a) method
int z = KillOver21(a); // This will return 0
System.out.print(z);
But you can achieve something like that with custom objects, say you have a class
class AHolder {
public int a;
}
then you can expect AHolder instance to change
public static void main(String [] args) {
AHolder a = new AHolder();
a.a = 34;
killOver21(a);
System.out.println(a.a);
}
public static void killOver21(AHolder b) {
if(b.a > 21) {
b.a = 0;
}
}
Since in the latter (even if its Pass by Value) , the reference is copied and both reference point to same object. So changes made inside the killOver21 method actually changes the object.
It is simply not possible, Java supports pass by value. int a's value will be copied to the function.
You could use Object instead of primitive where the reference value will be copied to your function by which you can get the actual object and modify it.
Fundamentally impossible in Java, period. int are immutable, and passed by value. You would need to create a mutable int type:
class MutableInt {
private int value;
public MutableInt(int value) { this.value = value; }
public getValue() { return this.value; }
public setValue(int value) { this.value = value; }
}
Then:
void KillOver21(MutableInt m) {
if(m.getValue() > 21) { m.setValue(0); }
}
However, be aware the mutable types that represent concepts that are defined by their value rather than their identity are generally an extremely bad idea. But, this is the only way to achieve what you're trying to achieve. Again, I caution you with the strongest words: what you're doing is a bad idea. You should find another way.
Doc, it hurts when I do this.
Then don't do that!
The simpliest way (quick&dirty) is to put value within an array
int holder[] = new int[]{ a};
KillOver21(holder)
System.out.printf( "value=[%d]", holder[0] );
void KillOver21(int holder[] ) {
holder[0] = 0;
}
I've got a question.
public class Jaba {
public static void main(String args[]) {
Integer i = new Integer(0);
new A(i);
System.out.println(i);
new B(i);
System.out.println(i);
int ii = 0;
new A(ii);
System.out.println(ii);
new B(ii);
System.out.println(ii);
}
}
class A {
public A(Integer i) { ++i; }
}
class B {
public B(int i) { ++i; }
}
To my mind passing an int\Integer as Integer to a function and making ++ on that reference should change the underlying object, but the output is 0 in all the cases. Why is that?
Most of the classes such as Integer that derive from Java's abstract Number class are immutable., i.e. once constructed, they can only ever contain that particular number.
A useful benefit of this is that it permits caching. If you call:
Integer i = Integer.valueOf(n);
for -128 <= n < 127 instead of:
Integer i = Integer.new(n)
you get back a cached object, rather than a new object. This saves memory and increases performance.
In the latter test case with a bare int argument, all you're seeing is how Java's variables are passed by value rather than by reference.
#Alnitak -> correct. And to add what really happens here. The ++i due to autoboxing works like that:
int val = Integer.intValue(); ++val;
and val is not stored anywhere, thus increment is lost.
As said in the other answers, Java does only call-by-value, and the ++ operator only effects a variable, not an object. If you want to simulate call-by-reference, you would need to pass a mutable object, like an array, and modify its elements.
The Java API has some specialized objects for this, like java.util.concurrent.atomic.AtomicInteger (which additionally also works over multiple threads), and org.omg.CORBA.IntHolder (used for call-by-reference for remote calls by the CORBA mechanism).
But you can also simply define your own mutable integer:
class MutableInteger {
public int value;
}
class C {
public C(int[] i) {
++i[0];
}
}
class D {
public D(MutableInteger i) {
++i.value;
}
}
class E {
public E(AtomicInteger i) {
i.incrementAndGet();
}
}
public class Jaba {
public static void main(String args[]) {
int[] iii = new int[]{ 0 };
System.out.println(iii[0]);
new C(iii);
System.out.println(iii[0]);
MutableInteger mi = new MutableInteger();
System.out.println(mi.value);
new D(mi);
System.out.println(mi.value);
MutableInteger ai = new AtomicInteger(0);
System.out.println(ai);
new E(ai);
System.out.println(ai);
}
}
If you want to use reference parameter then try this.
IntHolder
http://docs.oracle.com/javase/7/docs/api/org/omg/CORBA/IntHolder.html
i am using reflection to copy values of fields from object of Class A to object of Class B.
However method in A returns Number, where as setter in B requires Long. Is there any generic way i could set the value. As of now as expected i get illegalArgumentException: argument type mismatch
class a
{
Number value1;
Number value2;
public Number getValue1(){return value1;}
public Number getValue2(){return value2;}
}
class b
{
Double value1;
Long value2;
public void setValue1(Double value){this.value1 = value;}
public void setValue2(Long value){this.value2 = value;}
}
Not sure if my question is unclear.
You could do
b.setValue2(a.getValue2().longValue());
But if a.value2 isn't actually an integer (e.g. it's a Double with a fractional component) this will lose data.
Correspondingly
b.setValue1(a.getValue1().doubleValue());
Edit
Ok I think I've got a grasp on your situation. Here's a dirty way to go about what you want to do. Basically you need to have a transform method which will transform a Number into another Number based on a chosen class. That class you get from the Method itself. So it will be something like this:
public static void main(String[] args) throws Exception {
A a = new A();
a.setValue1(1.0);
a.setValue2(5);
B b = new B();
Method[] methods = b.getClass().getMethods();
for ( Method m : methods ) {
if ( m.getName().equals("setValue2") ) {
m.invoke(b, transform(a.getValue2(), m.getParameterTypes()[0]));
}
}
System.out.println(b.getValue2());
}
private static Number transform(Number n, Class<?> toClass) {
if ( toClass == Long.class ) {
return n.longValue();
} else if ( toClass == Double.class ) {
return n.doubleValue();
}
//instead of this you should handle the other cases exhaustively
return null;
}
The reason you would otherwise get an IllegalArgumentException in the above is because with a, value2 is not being set to a Long, it's being set to an Integer. They are disjoint types. If a.value2 was actually set to be a Long instead, you wouldn't have that error.
You need to do the conversion:
// get the Number 'number'
Long l = new Long(number.longValue());
// store the Long
You could do it even more efficiently using autoboxing.
You can't do this in a 'generic' way, because a Number could be a Float, Byte etc.
How can I cast an Object to an int in java?
If you're sure that this object is an Integer :
int i = (Integer) object;
Or, starting from Java 7, you can equivalently write:
int i = (int) object;
Beware, it can throw a ClassCastException if your object isn't an Integer and a NullPointerException if your object is null.
This way you assume that your Object is an Integer (the wrapped int) and you unbox it into an int.
int is a primitive so it can't be stored as an Object, the only way is to have an int considered/boxed as an Integer then stored as an Object.
If your object is a String, then you can use the Integer.valueOf() method to convert it into a simple int :
int i = Integer.valueOf((String) object);
It can throw a NumberFormatException if your object isn't really a String with an integer as content.
Resources :
Oracle.com - Autoboxing
Oracle.com - Primitive Data types
On the same topic :
Java: What's the difference between autoboxing and casting?
Autoboxing: So I can write: Integer i = 0; instead of: Integer i = new Integer(0);
Convert Object into primitive int
Scenario 1: simple case
If it's guaranteed that your object is an Integer, this is the simple way:
int x = (Integer)yourObject;
Scenario 2: any numerical object
In Java Integer, Long, BigInteger etc. all implement the Number interface which has a method named intValue. Any other custom types with a numerical aspect should also implement Number (for example: Age implements Number). So you can:
int x = ((Number)yourObject).intValue();
Scenario 3: parse numerical text
When you accept user input from command line (or text field etc.) you get it as a String. In this case you can use Integer.parseInt(String string):
String input = someBuffer.readLine();
int x = Integer.parseInt(input);
If you get input as Object, you can use (String)input, or, if it can have an other textual type, input.toString():
int x = Integer.parseInt(input.toString());
Scenario 4: identity hash
In Java there are no pointers. However Object has a pointer-like default implementation for hashCode(), which is directly available via System.identityHashCode(Object o). So you can:
int x = System.identityHashCode(yourObject);
Note that this is not a real pointer value. Objects' memory address can be changed by the JVM while their identity hashes are keeping. Also, two living objects can have the same identity hash.
You can also use object.hashCode(), but it can be type specific.
Scenario 5: unique index
In same cases you need a unique index for each object, like to auto incremented ID values in a database table (and unlike to identity hash which is not unique). A simple sample implementation for this:
class ObjectIndexer {
private int index = 0;
private Map<Object, Integer> map = new WeakHashMap<>();
// or:
// new WeakIdentityHashMap<>();
public int indexFor(Object object) {
if (map.containsKey(object)) {
return map.get(object);
} else {
index++;
map.put(object, index);
return index;
}
}
}
Usage:
ObjectIndexer indexer = new ObjectIndexer();
int x = indexer.indexFor(yourObject); // 1
int y = indexer.indexFor(new Object()); // 2
int z = indexer.indexFor(yourObject); // 1
Scenario 6: enum member
In Java enum members aren't integers but full featured objects (unlike C/C++, for example). Probably there is never a need to convert an enum object to int, however Java automatically associates an index number to each enum member. This index can be accessed via Enum.ordinal(), for example:
enum Foo { BAR, BAZ, QUX }
// ...
Object baz = Foo.BAZ;
int index = ((Enum)baz).ordinal(); // 1
Assuming the object is an Integer object, then you can do this:
int i = ((Integer) obj).intValue();
If the object isn't an Integer object, then you have to detect the type and convert it based on its type.
#Deprecated
public static int toInt(Object obj)
{
if (obj instanceof String)
{
return Integer.parseInt((String) obj);
} else if (obj instanceof Number)
{
return ((Number) obj).intValue();
} else
{
String toString = obj.toString();
if (toString.matches("-?\d+"))
{
return Integer.parseInt(toString);
}
throw new IllegalArgumentException("This Object doesn't represent an int");
}
}
As you can see, this isn't a very efficient way of doing it. You simply have to be sure of what kind of object you have. Then convert it to an int the right way.
You have to cast it to an Integer (int's wrapper class). You can then use Integer's intValue() method to obtain the inner int.
Answer:
int i = ( Integer ) yourObject;
If, your object is an integer already, it will run smoothly. ie:
Object yourObject = 1;
// cast here
or
Object yourObject = new Integer(1);
// cast here
etc.
If your object is anything else, you would need to convert it ( if possible ) to an int first:
String s = "1";
Object yourObject = Integer.parseInt(s);
// cast here
Or
String s = "1";
Object yourObject = Integer.valueOf( s );
// cast here
I use a one-liner when processing data from GSON:
int i = object != null ? Double.valueOf(object.toString()).intValue() : 0;
If the Object was originally been instantiated as an Integer, then you can downcast it to an int using the cast operator (Subtype).
Object object = new Integer(10);
int i = (Integer) object;
Note that this only works when you're using at least Java 1.5 with autoboxing feature, otherwise you have to declare i as Integer instead and then call intValue() on it.
But if it initially wasn't created as an Integer at all, then you can't downcast like that. It would result in a ClassCastException with the original classname in the message. If the object's toString() representation as obtained by String#valueOf() denotes a syntactically valid integer number (e.g. digits only, if necessary with a minus sign in front), then you can use Integer#valueOf() or new Integer() for this.
Object object = "10";
int i = Integer.valueOf(String.valueOf(object));
See also:
Inheritance and casting tutorial
int i = (Integer) object; //Type is Integer.
int i = Integer.parseInt((String)object); //Type is String.
Can't be done. An int is not an object, it's a primitive type. You can cast it to Integer, then get the int.
Integer i = (Integer) o; // throws ClassCastException if o.getClass() != Integer.class
int num = i; //Java 1.5 or higher
You can't. An int is not an Object.
Integer is an Object though, but I doubt that's what you mean.
If you mean cast a String to int, use Integer.valueOf("123").
You can't cast most other Objects to int though, because they wont have an int value. E.g. an XmlDocument has no int value.
I guess you're wondering why C or C++ lets you manipulate an object pointer like a number, but you can't manipulate an object reference in Java the same way.
Object references in Java aren't like pointers in C or C++... Pointers basically are integers and you can manipulate them like any other int. References are intentionally a more concrete abstraction and cannot be manipulated the way pointers can.
int[] getAdminIDList(String tableName, String attributeName, int value) throws SQLException {
ArrayList list = null;
Statement statement = conn.createStatement();
ResultSet result = statement.executeQuery("SELECT admin_id FROM " + tableName + " WHERE " + attributeName + "='" + value + "'");
while (result.next()) {
list.add(result.getInt(1));
}
statement.close();
int id[] = new int[list.size()];
for (int i = 0; i < id.length; i++) {
try {
id[i] = ((Integer) list.get(i)).intValue();
} catch(NullPointerException ne) {
} catch(ClassCastException ch) {}
}
return id;
}
// enter code here
This code shows why ArrayList is important and why we use it. Simply casting int from Object. May be its helpful.
For Example Object variable; hastaId
Object hastaId = session.getAttribute("hastaID");
For Example Cast an Object to an int,hastaID
int hastaID=Integer.parseInt(String.valueOf(hastaId));
Refer This code:
public class sample
{
public static void main(String[] args)
{
Object obj=new Object();
int a=10,b=0;
obj=a;
b=(int)obj;
System.out.println("Object="+obj+"\nB="+b);
}
}
so divide1=me.getValue()/2;
int divide1 = (Integer) me.getValue()/2;
We could cast an object to Integer in Java using below code.
int value = Integer.parseInt(object.toString());
If you want to convert string-object into integer...
you can simply pass as:
int id = Integer.valueOf((String) object_name);
Hope this will be helpful :-)
Integer x = 11
int y = x.intValue();
System.out.println("int value"+ y);
Finally, the best implementation for your specification was found.
public int tellMyNumber(Object any) {
return 42;
}
first check with instanceof keyword . if true then cast it.