I am very confused with the concept of using pointers in c

Question

I think I have misconceptions in my understanding on pointers?

Based on my understanding, int * twenties means that twenties is a pointer to an int .

So for *twenties = dollars/20; , the *twenties here refers to the value of the pointer?

In pay_amount(money, &twenties, &tens, &fives, &ones); , the pointer twenties is storing the address of &twenties in the function of pay_amount ? Wouldn't the twenties in printf("$20 bills: %d\n", twenties); print out the address instead of the value?

#include <stdio.h>

void pay_amount(int dollars, int *twenties, int *tens, int *fives, int *ones);

int main(void) {

    int money = 0, twenties, tens, fives, ones;

    printf("Enter a dollar amount: ");
    scanf("%d", &money);
    pay_amount(money, &twenties, &tens, &fives, &ones);

    printf("$20 bills: %d\n", twenties); // why isnt not *twenties??? I got an error if I put *twenties
    printf("$10 bills: %d\n", tens);
    printf(" $5 bills: %d\n", fives);
    printf(" $1 bills: %d\n", ones);

    return 0;
}

void pay_amount(int dollars, int *twenties, int *tens, int *fives, int *ones) {

    *twenties = dollars / 20;
    dollars -= *twenties * 20;
    *tens = dollars / 10;
    dollars -= *tens * 10;
    *fives = dollars / 5;
    dollars -= *fives * 5;
    *ones = dollars;
}

Answer 1

No, the twenties in the main() and the twenties in the pay_amount function are different types.

While calling the pay_amount() from main() , you're passing the address of twenties of main() , and storing that in twenties (which is local to the block scope of the function) of pay_amount .

The type of twenties in main() is int , the type of twenties in pay_amount is int * .

If you wish, you can use two different variable names altogether.

So, in main() , twenties is an int , and a print statement like

 printf("$20 bills: %d\n", twenties);

is correct. Point to note, if you want to print the twenties inside the pay_amount() function call, you have to use *twenties , as you'd have expected, as that one is of type int * .

Answer 2

So, if you declare two variables as:

int data_holder;
int* pointer;

And assign values to them as:

data_holder = 10;
pointer = &data_holder;

What happens is this - compiler allocates memory for data_holder as well as pointer . Now, when you store 10 in data_holder and use & to get its address and store it in pointer , a link is created. To access the value 10 - you may either use data_holder or de-reference pointer .

EDIT : When we use * during declaration of a pointer variable, it tells the compiler to treat the variable differently. When we use * again for that declared pointer, we are essentially de-referencing the pointer - accessing the value stored at the memory location it is pointing to.

More concretely;

printf("The value of data_holder is %d", data_holder);
printf("The value of data_holder is %d", *pointer);

Similarly, the value 10 can be modified in two ways;

data_holder = 200;       //The value is changed from 10 to 200
*pointer = 300;          //The value is changed from 200 to 300

Inside your function, twenties etc. are pointers, whereas in main() the twenties etc. are integers.