Can we swap two numbers in Java using pass by reference or call by reference?
Recently when I came across swapping two numbers in Java I wrote
class Swap{
int a,b;
void getNos(){
System.out.println("input nos");
a = scan.nextInt();
b = scan.nextInt(); // where scan is object of scanner class
}
void swap(){
int temp;
temp = this.a;
this.a = thisb;
this.b = this.a;
}
}
In the main method I call the above mentioned methods and take two integers a,b and then using the second method I swap the two numbers, but relative to the object itself....
Does this program or logic come under pass by reference?
And is this correct solution?
Yes and no. Java never passes by reference, and your way is one workaround. But yet you create a class just to swap two integers. Instead, you can create an int wrapper and use pass it, this way the integer may be separated when not needed:
public class IntWrapper {
public int value;
}
// Somewhere else
public void swap(IntWrapper a, IntWrapper b) {
int temp = a.value;
a.value = b.value;
b.value = temp;
}
As the comments show, I might not have been clear enough, so let me elaborate a little bit.
What does passing by reference mean? It means that when you pass an argument to the method, you can change the original argument itself inside this method.
For example, if Java was pass-by-reference, the following code will print out x = 1:
public class Example {
private static void bar(int y) {
y = 10;
}
public static void main(String[] args) {
int x = 1;
bar(x);
System.out.println("x = " + x);
}
}
But as we know, it prints 0, since the argument passed to the bar method is a copy of the original x, and any assignment to it will not affect x.
The same goes with the following C program:
static void bar(int y) {
y = 1;
}
int main(int argc, char * argc[]) {
int x = 0;
bar(x);
printf("x = %d\n", x);
}
If we want to change the value of x, we will have to pass its reference (address), as in the following example, but even in this case, we will not pass the actual reference, but a copy of the reference, and by dereferencing it we will be able to modify the actual value of x. Yet, direct assignment to the reference will no change the reference itself, as it is passed by value:
static void bar(int &y) {
*y = 1;
y = NULL;
}
int main(int argc, char * argc[]) {
int x = 0;
int * px = &x;
bar(px);
printf("x = %d\n", x); // now it will print 1
printf("px = %p\n", px); // this will still print the original address of x, not 0
}
So passing the address of the variable instead of the variable itself solves the problem in C. But in Java, since we don't have addresses, we need to wrap the variable when we want to assign to it. In case of only modifying the object, we don't have that problem, but again, if we want to assign to it, we have to wrap it, as in the first example. This apply not only for primitive, but also for objects, including those wrapper objects I've just mentioned. I will show it in one (longer) example:
public class Wrapper {
int value;
private static changeValue(Wrapper w) {
w.value = 1;
}
private static assignWrapper(Wrapper w) {
w = new Wrapper();
w.value = 2;
}
public static void main(String[] args) {
Wrapper wrapper = new Wrapper();
wrapper.value = 0;
changeValue(wrapper);
System.out.println("wrapper.value = " + wrapper.value);
// will print wrapper.value = 1
assignWrapper(w);
System.out.println("wrapper.value = " + wrapper.value);
// will still print wrapper.value = 1
}
}
Well, that's it, I hope I made it clear (and didn't make too much mistakes)
import java.util.*;
public class Main
{
int a,b;
void swap(Main ob)
{
int tmp=ob.a;
ob.a=ob.b;
ob.b=tmp;
}
void get()
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter a and b: ");
a=sc.nextInt();
b=sc.nextInt();
}
public static void main(String[] args) {
Main ob=new Main();
ob.get();
ob.swap(ob);
System.out.println(ob.a+" "+ob.b);
}}
Related
I'm working on a calculator and I search how I can optimize my code.
The thing is that I have much code duplication due to if I'm working on the first number of the calculation or the second. So I'm searching if it is possible to modify the value of an attribute sent in argument of a function ? (I think not because I saw nowhere the answer).
Maybe I'm expressing myself badly so here is a code below to explain what I'm talking about:
public class MyClass
{
private static int number1 = 1;
private static int number2 = 2;
public MyClass()
{
changeValueOf(number1, 3);
}
private static void changeValueOf(int number, int value)
{
//Change here the value of the correct field
}
}
First of all, you can modify static variables inside the method:
private static void changeValueOf(int value)
{
number1 = value;
}
But I guess that is not what you a looking for :)
In Java (and in most other languages) primitive data type (int, short, long, etc) passed by value, e.g. the copy of value passes to the method (function).
And reference types (objects, e.g. created with new operator) passed by reference. So, when you modigy the value of reference type (object) you can see the changes in the outer scopes (for example, in method caller).
So, the answer is no - you cannot change the value of int so that the outer scope would see the updated value.
Howewer, you could wrap your int values with some object - and it change the value inside of it:
public class Example {
public static void main(String[] args) {
Example app = new Example();
// Could be static as well
Holder val1 = new Holder(1);
Holder val2 = new Holder(2);
app.changeValue(val1, 7);
System.out.println(val1.value); // 7
}
public void changeValue(Holder holder, int newValue) {
holder.value = newValue;
}
static class Holder {
int value;
Holder(int value) {
this.value = value;
}
}
}
Also, you could create an array with 2 values and update them inside the method, but it's not very good approach IMO
And finally, you could just return updated value and assign it to your variables:
public class Example {
private static int number1 = 2;
private static int number2 = 3;
public static void main(String[] args) {
Example app = new Example();
number1 = app.mul(number1, 7);
number2 = app.mul(number2, 7);
System.out.println(number1); // 14
System.out.println(number2); // 21
}
public int mul(int a, int b) {
return a * b;
}
}
One possibility is to use an array to store your variables, instead of separate variables with numbers affixed. Then you would write number[1] instead of number1 for example. You can pass the array index number around to indicate which variable you are referring to.
public class MyClass
{
private static int[] variables = {1, 2};
public MyClass()
{
// change value of first variable
changeValueOf(0, 3);
// now variable[0] = 3
}
private static void changeValueOf(int number, int value)
{
variables[number] = value;
}
}
#include<stdio.h>
void decrease(int *i);
int main(){
int i = 10;
decrease(&i);
printf("%d",i);
}
void decrease(int *i){
*i = *i - 1;
}
What would be the Java program for the same?
As you pointed out (no pun intended), Java does not support pointers. So, there is no way to directly manipulate the value of a primitive passed to a method, because only a copy of the primitive would be used in the method. One way to get around this would be to just return the updated value, and then overwrite the integer in the calling scope:
public static int decrease(int i) {
return i - 1;
}
public static void main(String[] args) {
int i = 10;
i = decrease(i);
System.out.println(i); // prints 9
}
You have two options, either (return) the value, and modify it in the main class, or pass an Object, not a primitive.
An object with your values:
public class Holder {
public int x;
}
And a method to modify it
public void modify(Holder h){
h.x = 2;
}
Called like:
Holder h = new Holder();
h.x = 1;
modify(h);
System.out.println(h.x);
Results in:
2
How do i print the value of variable which is defined inside another method?
This might be a dumb question but please help me out as i am just a beginner in programming
public class XVariable {
int c = 10; //instance variable
void read() {
int b = 5;
//System.out.println(b);
}
public static void main(String[] args) {
XVariable d = new XVariable();
System.out.println(d.c);
System.out.println("How to print value of b here? ");
//d.read();
}
}
You can't. b is a local variable. It only exists while read is executing, and if read executes multiple times (e.g. in multiple threads, or via recursive calls) each execution of read has its own separate variable.
You might want to consider returning the value from the method, or potentially using a field instead - it depends on what your real-world use case is.
The Java tutorial section on variables has more information on the various kinds of variables.
You need to return value from your read() methods.
public class XVariable {
int c = 10; //instance variable
int read() {
int b = 5;
return b;
}
public static void main(String[] args) {
XVariable d = new XVariable();
System.out.println(d.c);
System.out.println(read());
//d.read();
}
}
Return b from the read method and print it
public class XVariable {
int c = 10; //instance variable
int read() {
int b = 5;
return b;
}
public static void main(String[] args) {
XVariable d = new XVariable();
System.out.println(d.c);
System.out.println(d.read());
}
}
I made this code to learn my self about how return works in java
public class test {
public int sumDouble(int a, int b) {
int k = (a + b);
if (a == b) {
k = k * 2;
} else {
k = (a + b);
}
return k;
}
public static void main(String[] args) {
System.out.println("Enter your number");
Scanner scan = new Scanner(System.in);
int a = scan.nextInt();
int b = scan.nextInt();
test t = new test();
t.sumDouble(a, b);
}
}
I want to use the return k to print out the value of summation
How can use the return k to print out the sum value?
I tried to write System.out.println(t.k); in the main method but it did not work ..
thanks
The thing to keep in mind here is that k is a local variable in the sumDouble method, so you can't refer to it elsewhere.
You can take the value returned by sumDouble and use it in your main method, however.
For instance, you can create a variable in main and assign its value to be the value that is returned from sumDouble, and you could print that variable, since it's in scope:
// at runtime, this will evaluate "t.sumDouble(a, b)"
// and assign the value that it returns to the variable "sum"
int sum = t.sumDouble(a, b);
System.out.println(sum);
You could even skip the local variable and use the method call directly:
System.out.println(t.sumDouble(a, b));
I'm trying to write a program that takes the output of adding two numbers in one class together and adds it to a different number. Here is the first class:
public class Add{
public static void main(String[] args) {
int a = 5;
int b = 5;
int c = a + b;
System.out.println(c);
}
}
And the second:
public class AddExtra{
public static void main(String[] args) {
Add a = new Add();
int b = 5;
int c = a.value+b;
System.out.println(c);
}
}
How do I get this to work? Thanks.
Suggestions:
You need to give the Add class a public add(...) method,
have this method accept an int parameter,
have it add a constant int to the int passed in,
and then have it return the sum.
If you want it to add two numbers, rather than a number and a constant, then give the method two int parameters, and add them together in the method.
Then create another class,
In this other class you can create an Add instance,
call the add(myInt) method,
and print the result returned.
You could try
public class Add{
public int c; // public variable
public Add() { // This is a constructor
// It will run every time you type "new Add()"
int a = 5;
int b = 5;
c = a + b;
}
}
Then, you can do this:
public class AddExtra{
public static void main(String[] args) {
Add a = new Add(); // Here, the constructor is run
int b = 5;
int c = a.c + b; // Access "a.c" because "c" is a public variable now
System.out.println(c);
}
}
Read more about constructors here.