Thread: My very first program...ever

I took the advice of some of you and googled college level homework assignments for c programming and was able to locate a few sources. Here is my very first program, written on my own, using my own logic and without any help.

Before you tear me up too much, please realize that I am sure this could be written much smaller and more efficiently. Your input is what I can use to learn not only to work with the logic of C, but the beginning of thinking on how to write more efficient code and how to write code in the most commonly accepted format.

If you want to lead me to a more compact design for this program, please try to hint your way around it so I can try and figure out what you mean. It will help me to gain a deeper understanding of the logic than if you come right out and tell me how to do something. Plus, if I get stumped...I will ask outright. LOL.

Thanks and double thanks for bearing with a newbie whom is trying to teach himself how to program.

Code:

/* develop a program that accepts a dollar amount (in cents) and then computes
the number of quarters, dimes, nickels, and pennies needed.  The solution must
produce the smallest number of coins possible for the given dollar amount*/


#include <stdio.h>

main()
{

    int c,q = 0, d = 0, n = 0, p = 0;     /* I had to initialize my actual coin 
                                          counts to zero, else I got some pretty
                                          amazing numbers*/
    
    printf( "Please enter an amount ( in cents ):" );
    scanf( "&#37;d", &c );
    getchar();
    
    while ( c >= 25 ){
          q++;
          c = c - 25;                               /*Could I have used c -= 25 for this and following statements*/
          }
    printf( "* You will need %d quarters.\n", q);
              
    while ( c >= 10 ){
          d++;
          c = c - 10;
          }
    printf( "* You will need %d dimes.\n", d);
    
    while ( c >= 5 ){
          n++;
          c = c - 5;
          }
    printf( "* You will need %d nickels.\n", n);
    
    while ( c >= 1 ){
          p++;
          c = c - 1;
          }
    printf( "* You will need %d pennies.\n", p);
    
    getchar();
}

That is certainly one way to do it. Another option is to use division and the modulus operator.

q = c / 25;
c = c % 25;

The first equation will give you the number of times the coin amount goes into c (this will be rounded down for you if you are using ints, so 3.14 would be 3). Example:

147 / 25 = 5.88 which will round down to 5

The second equation uses the modulus operator which will return the remainder of a division. So:

147 % 25 = 22

This accomplishes the same thing as your while loops. If you want to still use the while loops you could move the logic into a function. Notice that you are doing the same thing for each coin with only the coin value changing. That is prime behavior that could be moved into a function. You might try this when you feel more comfortable with the basics.

Looks good for your first program. Have fun!

> Could I have used c -= 25 for this and following statements

Yes.

I thought there was a way to put all of my loops into a single function to better serve what I was trying to accomplish, but I'm just beginning to read about functions and how they work, so that could take some time for me to work out.

What exactly is difference between main(), void main(), and int main()?

Also, what effect does return 0 have on a program and how does its absence affect the same?

I didn't even think of using modulus. Thanks for that one.

Okay thanks.

Here is an example of how you could do this a little differently and reuse your while approach. This introduces functions and pointers (although you have already used a pointer with scanf).

Code:

#include <stdio.h>

int payChange(int* change, int coin); /* function prototype */

int main()
{

    int change;

    printf( "Please enter an amount ( in cents ):" );
    scanf( "%d", &change);
    getchar();

    printf( "* You will need %d quarters.\n", payChange(&change, 25));
    printf( "* You will need %d dimes.\n", payChange(&change, 10));
    printf( "* You will need %d nickels.\n", payChange(&change, 5));
    printf( "* You will need %d pennies.\n", payChange(&change, 1));
    
    getchar();

    return(0);
}

/* takes in a pointer to the change to be paid and the coin to use to pay it */
int payChange(int* change, int coin)
{
	int amount = 0;
	while (*change >= coin)
	{
		amount++;
		*change = *change - coin;
	}
	return(amount);
}

Look into pointers and see if this makes sense. Maybe you have already looked into this if you wondered why scanf needed an & in front of the variable.

Thanks, I'll toy around with it and see how it goes. Another question. If I wanted my output to only print lines that had a coin count higher than zero, would I be best to try to make a function or to use conditional statements? Either way, please don't show me the code...just kind of *point me in the right direction..thanks.

It's good to know there are those that are interested in helping people out there. Thats a rarity found in life these days.

Originally Posted by Dogmasur

Thanks, I'll toy around with it and see how it goes. Another question. If I wanted my output to only print lines that had a coin count higher than zero, would I be best to try to make a function or to use conditional statements? Either way, please don't show me the code...just kind of *point me in the right direction..thanks.

It's good to know there are those that are interested in helping people out there. Thats a rarity found in life these days.

An if statment would work well. Try to make your program as simple as possible. Ask yourself, what if they suddenly discontinued dimes? Would I need to edit 10 different lines of my program? Or could I just change one line slightly? What if they added a 50 cent piece?

One thing you might try is modifying your coin program to require the smallest modification to support a new type of coin.

I rewrote my code to match yours above and it worked fine. I asked about using a conditional expression to modify my program so that it will not print lines that end up having a coin count of zero. You made some suggestions and I have been toying around for a day and have to admit I am completely stumped. I don't see it. Can you show it to me and explain the process.

Thanks for your time.

Originally Posted by Dogmasur

I rewrote my code to match yours above and it worked fine. I asked about using a conditional expression to modify my program so that it will not print lines that end up having a coin count of zero. You made some suggestions and I have been toying around for a day and have to admit I am completely stumped. I don't see it. Can you show it to me and explain the process.

Thanks for your time.

You'd probably need to use some variables in this case.

Code:

int quarters;
quarters = payChange(&change, 25);
if (quarters > 0)
     printf(" * You will need %d quarters.\n", quarters);

Did this answer your question?

Actually I thought of something similar to this. Would I then need to use an if for each coin iteration? Your last reply made it seem like I should try to use as little code as possible, so I was thinking of a way of trying to use one code block ( or maybe a function particular to this operation ) to handle all of my printf iterations. Am I not understanding your most recent reply or am I reading too much into it?

Thanks

That makes some sense. I read a little bit about structures, and have to admit that I think those are beyond me yet. I'll play around with the array idea for a bit and see if I can figure this out.

Thanks for the help.