Consider the following C functionvoid swap (int a, int b){    int temp;    temp = a;     a = b;     b = temp;}.In order to exchange the values of two variables x and y.a.Call swap (x, y)b.Call swap (&x, &y)c.swap(x,y) cannot be used as it does not return any valued.swap(x,y) cannot be used as the parameters are passed by value

Given datatype of two variable as an integer value and address of two variables as a void pointer to the function common_swap(). Do swap the values pointed by the two pointer variables.  The integer value from 1 to 6 means the datatype. 1. char(%c) 2. short int(%hd) 3.long int(%ld) 4. long long int(%lld) 5. float(%f) 6. double(%lf).

Using pointers write a program to swap two numbers using a user definedfunction(Call by reference)

Consider the following method that is intended to swap the values of two integers:1234567891011public static void falseSwap(int a, int b) {    int temp = a;    a = b;    b = temp;}public static void main(String[] args) {    int x = 3;    int y = 4;    falseSwap(x, y);    System.out.println(x + " " + y);}Why doesn’t the falseSwap method swap the contents of x and y?

What is the purpose of the swap function in this program?*1 pointA. To swap the positions of two adjacent elements in the array.B. To swap the values of two variables in the main function.C. To swap the elements at the beginning and end of the array.D. To swap the positions of every two elements in the array.After the fourth pass, what is the array state?*1 pointA. {1, 2, 4, 3, 5, 6, 7, 8, 9, 10}B. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}C. {2, 1, 4, 5, 7, 3, 8, 6, 9, 10}D. {10, 9, 8, 7, 6, 5, 4, 3, 2, 1}

12345678910111213141516#include <stdio.h>void swap(int *x, int *y) { int temp = *x; *x = *y; *y = temp; printf("%d %d", *x, *y);}int main() { int a = 120; int b = 121; printf("%d %d\n", a, b); swap(&a, &b); return 0;}