Scenario : You have two variables a and b with values assigned to them as 100 and 101 respectively. Write a program in java using third variable two swap the values within the a and b.
Input : a = 100 and b = 101
Output : a = 101 and b = 100
public class swap {
public static void main(String[] args)
{
int a = 100;
int b = 101;
System.out.println("Before
swapping, a :"+ a +" and b : "+ b);
swapValue(a,
b);
}
public static void swapValue(int a, int b){ //mind the use of static here*
int temp; //temp is the third variable
temp
= a;
a
= b;
b
= temp;
System.out.println("After
swapping, a : "+ a +" and b : "+ b);
}
}
OUTPUT :
Before swapping, a : 100 and b : 101
After swapping, a : 101 and b : 100
mind the use of static here*
Try removing static from this line, public static void swapValue(int a, int b) and see what you get.
You will get compile-time error at swapValue(a, b); saying "Cannot make a static reference to the non-static method swapValue(int, int) from the type swap".
Why so? (Learn here)
Don't sit relaxed!
Don't sit relaxed!
Do you really think the original value of variable a and b has been swapped?
To check it, do just as I instruct. Try to print the value of a and b after the swapValue(a, b) method complete its task.
public class swap {
public static void main(String[] args) {
int a = 100;
int b = 101;
System.out.println("Before swapping, a :"+ a +" and b : "+ b);
swapValue(a, b);
System.out.println("Is swapping done? a :"+ a +" and b : "+ b);
System.out.println("Is swapping done? a :"+ a +" and b : "+ b);
}
public static void swapValue(int a, int b){ //mind the use of static here*
int temp; //temp is the third variable
temp = a;
a = b;
b = temp;
System.out.println("After swapping, a : "+ a +" and b : "+ b);
}
}
And surprisingly you will see the println in red would print the original value of a and b. i.e. a = 100 and b = 101
----------------------------------------------------------------------------------------
Even though swapValue() method shows that the value of the two variable is swapped then why this crap???
Reason is pretty simple... Look at these two lines again,
swapValue(a, b); //a, b are the variables declared in main() method
public static void swapValue(int a, int b) //here a, b are the local variable of this // swapValue() method.
public static void swapValue(int c, int d)
Now what is happening in above line. The value a, b passed by swapValue(a, b) from the main() method is copied in the local variable c(will hold value of a) and d (will hold value of b) of
public static void swapValue(int c, int d){
int temp; //temp is the third variable
temp = c;
c = d;
d = temp;
System.out.println("After swapping, c : "+ c +" and d : "+ d);
}
So the logic written inside this method is actually swapping c, d which is local variable of this method containing copied value of a, b passed in calling function swapValue(a, b) from main().
Thus we can see it is not the value of a, b passed in the calling function from main() method is swapped but local variable in called function is getting swapped.
Hope I made it clear to you...
If not please ask me again
