Swapping variables values in C using Pointers, C (gcc)

Swapping variables values in C using Pointers

// Original program name: Change Variable Without Third Variable

#include  <stdio.h>

// Here x, and y are pointers. Which means that I am directly affecting the value present in the variables' address, and I don't even need to
// write a return statement.
void swapWithoutThirdVariable(int *x, int *y) {
    *x = *x + *y;
    *y = *x - *y;
    *x = *x - *y;
}

// Here x, and y are pointers. Which means that I am directly affecting the value present in the variables' address, and I don't even need to
// write a return statement.
void swapWithThirdVariable(int *x, int *y) {
    int temp = *x;
    *x = *y;
    *y = temp;
}

int main(void) {
    int X = 21, Y = 632;
    printf("%d\t%d\n", X, Y);
    
    // I am passing the addresses of the variables of X, and y. 
    //Now, I will manipulate the values present in the addresses themselves, that will directly affect the variables themselves.
    swapWithoutThirdVariable(&X, &Y); 
    printf("%d\t%d\n", x, y); // Printing the swapped values.
    
    swapWithThirdVariable(&X, &Y); // Swapping them again, so X gets back it's original value, and same with Y.
    printf("%d\t%d", x, y);
    return 0;
}
// THIS SHIT WAS POSTED 8 YEARS AGO. I BROUGHT THIS RELIC BACK.
// I don't know who originally made this.
// PROGRAM EDITED BY 'JAMES COLLINS'


run \| edit \| history \| help	1

Please log in to post a comment.