Someone explain this in simple way ..y the answer is 5 not 6?

Question

public class MyClass {
public static void main(String[ ] args) {
int x = 5;
addOneTo(x);
System.out.println(x);
}

static void addOneTo(int num) {
num = num + 1;
}
}

Answer 1

wait what was that john???
u r using * and & in java??
java has no such conventions
-_- -_-

Answer 2

it will work in c++ right ?

Answer 3

public class MyClass {
public static void main(String[ ] args) {
int x = 5;
int c;
c=addOneTo(x);
System.out.println(c);
}

static int addOneTo(int num) {
num = num + 1;
return num;
}
}

addoneTo function needs to return the incrememted value...
else num the change of x wont occur in main func...

if u want to do in either
u need to tae x as global var..

Answer 4

Pointers are not supported in Java, so @John's Idea won't work.

Answer 5

ya cpp has pointer...
that ans has been marked as best ans...😨😨😵😵OMG

Answer 6

Anyways, you could also try this:

public class MyClass {
static int x;
public static void main(String[ ] args) {
x = 5;
addToX ();
System.out.println(x);
}
static void addToX (){
x++;
}
}

The function does not work with other variables as x here, so @sayans idea might be better for your uses.

Answer 7

I posted the same code as yours @sayan

Answer 8

Ok, your point😊

Answer 9

this is another way...

public class MyClass
{
static int x;
public static void main(String[ ] args)
{
x = 5;
addToX ();
System.out.println(x);
}
static void addToX ()
{
x++;
}
}

Answer 10

no you r not...
1--the code needs a static before int...
ur code wont work...
2--also u have printed c
which is not in code..