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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;}

Question

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;}

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Solution

The text you provided is a C program. It defines a function swap that swaps the values of two integers, and a main function that creates two integers a and b, prints their values, swaps their values using the swap function, and then ends.

Here's a step-by-step explanation:

  1. The main function is the entry point of the program. It starts by declaring two integer variables a and b and assigning them the values 120 and 121, respectively.

  2. It then prints the values of a and b using the printf function. The %d in the printf function is a placeholder for an integer, so it prints the values of a and b.

  3. The swap function is then called with the addresses of a and b as arguments. The & operator is used to get the address of a variable.

  4. Inside the swap function, a temporary variable temp is declared and assigned the value of *x, which is the value at the address of a.

  5. The value at the address of x is then changed to the value at the address of y, effectively changing the value of a to the value of b.

  6. The value at the address of y is then changed to temp, effectively changing the value of b to the original value of a.

  7. The swap function then prints the new values of *x and *y, which are the swapped values of a and b.

  8. The main function then returns 0, signaling that the program has ended successfully.

This problem has been solved

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123456789101112131415161718#include <stdio.h>int f(int x) { if (x == 2) { return 2; } else { printf("*"); return f(x - 1); }}int main() { int n, i; n = f(6); printf("%d", n); return 0;}

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