/**
 * print_addresses.c
 *
 * Illustrates argument passing in C with memory addresses.
 *
 * Compile:  Using Linux and gcc
 *           gcc -g -Wall -o print_addresses print_addresses.c
 * Run:      ./print_addresses
 */

#include <stdio.h>

void by_value(int number);
void by_reference(int * number_p);

int main(int argc, char *argv[]) {
    int num1;

    printf("Enter an integer: ");
    scanf("%d", &num1);

    printf("[ 1 ] Number = [ %d ], and its address is [ %p ]\n", num1, &num1);

    by_value(num1);
    by_reference(&num1);

    int num2 = num1;

    printf("[ 2 ] Number = [ %d ], and its address is [ %p ]\n", num2, &num2);
    by_value(num2);
    by_reference(&num2);

    return 0;
}

/* Here, we are passing in a *value*. In other words, our 'actual args'
 * contain the integer itself. Because of this a copy is being made in memory,
 * and will have a different memory address. */
void by_value(int number) {
    printf("[VAL] Number = [ %d ], and its address is [ %p ]\n", number, &number);
}

/* Here, we are passing in a *pointer*, something that refers to a memory
 * address. No copy of value takes place; note how the address will be the same
 * as the address of what was passed in. */
void by_reference(int * number_p) {
    printf("[REF] Number = [ %d ], and its address is [ %p ]\n", *number_p, number_p);
}