C program

Hello,

I am learning C right now...anyway I do try to understand it but I got a problem with the following program. It's a C program that calculates the exact change. So, since only int arithmetic is exact, I l need to break up your double into two ints, the one before the decimal point and the one after the decimal point. It gotta look like this here

Enter the sales price: 5.66
Enter the cash paid: 10.75
Here's your change: 5.09
5 dollars, 1 nickel and 4 pennies
Want to try again? [y/n] n

I have like no idea now how do do this one. Can someone help me, please? Very much appreciated!!

You have no need to do that in the first place. Use double or float instead of int.

Quote:

Originally Posted by Elysia

You have no need to do that in the first place. Use double or float instead of int.

Yes, I thought the same. But I am still not able to do it :(

char buf[50];
double price;
fgets(buf, sizeof(buf), stdin);
price = strtod(buf, NULL, 10);

If you insist upon using the two int approach, this is a somewhat elaborate way to do it:

Create a struct to represent an amount, containing two ints, one for dollars and one for cents. Create functions to do the necessary arithmetic on such structs. When you want to read an amount, you can use scanf() or fgets()/sscanf() with the format specificer of something along the lines of "%d.%d".

If you do that, the rest of the program might be trivial.

Another way to think of it is to change all your amounts into pennies by multiplying them X 100.

Now 10.57 becomes 1057, and 5.95 becomes 595. Everything becomes an int since there are no tenths or hundred's of a penny.

To change it back to normal change, use a while loop:

Code:

dollars = quarters = dimes = nickels = pennies = 0; while(amount > 0) { if(amount > 100) amount = amount - 100; dollars++; } else if(amount > 24) { amount = amount - 25; quarters++; } //continue with this logic with dimes, nickels, and pennies, until you have diminished //amount to zero. } //end of while loop

Note that if you want the highest change, you'd want to use something like this:

Code:

while(amount >= 100) { amount -= 100; dollars ++; } /* ... */

(Note the >= instead of >, as well.)

Quote:

Originally Posted by Adak

Another way to think of it is to change all your amounts into pennies by multiplying them X 100.

Now 10.57 becomes 1057, and 5.95 becomes 595. Everything becomes an int since there are no tenths or hundred's of a penny.

To change it back to normal change, use a while loop:

Code:

dollars = quarters = dimes = nickels = pennies = 0; while(amount > 0) { if(amount > 100) amount = amount - 100; dollars++; } else if(amount > 24) { amount = amount - 25; quarters++; } //continue with this logic with dimes, nickels, and pennies, until you have diminished //amount to zero. } //end of while loop

So basically now I have to add the fields for the input and output and Im done?

Do you know how to write a number in binary? You start with the highest (most significant) digit. For example, to write 10 in binary, you'd see that there was an 8, and so put a 1 in the 8s column.

Code:

1???

Now you're left with 10-8 = 2. Consider the next column, the fours column: there is no 4 in 2.

Code:

10??

But there is a 2 in 2.

Code:

101?

Which leaves you with 0. There aren't any ones in 0.

Code:

1010

It's the same basic idea to convert a numerical amount into change. Start with the largest denomination -- say, whole dollars. Use as many of those as you can -- you'll eventually end up with a number less than a dollar. Now go on to the next smallest change -- say, quarters. Use as many quarters as you can, until you have less than 25 cents, and so on.

It's just like the code I posted. I'll extend it a little.

Code:

while(amount >= 100) { amount -= 100; dollars ++; } while(amount >= 25) { amount -= 25; quarters ++; } /* ... */

Of course, you could make your life easier and use / and %.

Code:

dollars = amount / 100; amount %= 100; quarters = amount / 25; amount %= 25; /* ... */

Little tidbit to consider... If you need to worry about printing plural or not, you should try the tertiary command.

Code:

: : if(dollars) (dollars > 1) ? printf("%d dollars ", dollars) : printf("%d dollar ", dollars); : : if(quarter) ... if(dime) ...

And so on till you reach your pennies.

And many such examples can be contrived:

Code:

printf("%d dollar%s", dollars, dollars == 1 ? "" : "s");

Quote:

Originally Posted by XodoX

So basically now I have to add the fields for the input and output and Im done?

No, you're a student, and I wouldn't let you go through this and be dumbed down.

You need to add your input block of code, extend the coin logic I showed before, all the way down to pennies, and add your output block of code.

As is mentioned above, and shown most eloquently by MacGyver, the tertiary operator is a great little addition to your code. That way "1 pennies" is never printed, becoming "1 penny", and "1 dollars" becomes "1 dollar", etc.

As much as anything else though, understand how the logic works here - especially WHY you have to start with the dollars in the while loop, instead of starting with the dimes, or quarters, etc. Also, why you need to use else if in the if statement, instead of a simpler else.

Notice also, that there are a myriad of ways to solve this problem. Some are simple and easy for nearly anyone to understand, others are complex and leave your imagination panting to catch it's breath. Others are somewhere in between.

Any of the solutions might be "the best", depending on the exact situation. Your goal imo is to find the solution that is most clear when you look at the problem to be solved, and at the code.

All right then...thank you.

okay at least I think I got a part of it now...

Code:

double SalesPrice = 5.66; double CashPaid = 10.75; int SPDollars; // Sales price dollars int SPCents; // Sales price cents int CPDollars; // Cash paid dollars int CPCents; // Cash paid cents int Dollars; // Number of $1 bills int Quarters; // Number of Quarters int Dimes; // Number of Dimes int Nickels; // Number of Nickels int Pennies; // Number of Pennies int Remaining; SPDollars = (int)SalesPrice; // SPDollars = 5 SPCents = (int)(SalesPrice * 100.00) % 100; // 5.66 * 100.00 = 566.00; 566 % 100 = 66 cents CPDollars = (int)CashPaid; CPCents = (int)(CashPaid * 100.00) % 100; Dollars = CPDollars - SPDollars; // 5 Remaining = CPCents - SPCents; // 9 if (Remaining < 0) { // We had to borrow a dollar Dollars--; Remaining += 100; }; Quarters = Remaining / 25; // How many quarters? 0 Remaining = Remaining % 25; // How much left? 9 Dimes = Remaining / 10; // How many dimes? 0 Remaining = Remaining % 10; // How much left? 9 Nickels = Remaining / 5; // How many nickels? 1 Pennies = Remaining % 5; // How many pennies? 4

Does that look good so far ? I know I still need to add the other part.

Overall it looks - a little over complicated (what do you care about sales price dollars and cents? Or cash price dollars and cash price cents?). You seem to have the math all set - good! :)

Half those variables should wing their way to the variable graveyard. They just muck up the program with unneeded junk, imo.

Run-time and Clarity are the clarion call for programs, and your program has no problem with run-time.

Quote:

Originally Posted by Adak

Overall it looks - a little over complicated (what do you care about sales price dollars and cents? Or cash price dollars and cash price cents?). You seem to have the math all set - good! :)

Half those variables should wing their way to the variable graveyard. They just muck up the program with unneeded junk, imo.

Run-time and Clarity are the clarion call for programs, and your program has no problem with run-time.

If I want to use any value for the input is it ok if I just do it like this

Code:

double SalesPrice = '\n; double CashPaid = '\n;

I'm not sure if this would be okay. Guess I have to delete the numbers after the "remaining".

Quote:

Originally Posted by XodoX

If I want to use any value for the input is it ok if I just do it like this

Code:

double SalesPrice = '\n; double CashPaid = '\n;

I'm not sure if this would be okay. Guess I have to delete the numbers after the "remaining".

Doubles are numbers, and '\n' (note the second single quote btw), is a newline char - which is just an 8 bit number, itself.

So, no.

double SalesPrice; (leaving SalesPrice uninitialized), is fine, or
double SalesPrice = 0.0; is fine

Either method above, is fine.

Thank you:)

It says thoug "want to try again?"

should I just make it reset ? If I just use printf to say yes or no it might be to simple. I dont know. Also dont know how I would do it though. Couldn't find a good C code database online.

Edit:

I gues I could do it with if. If xx then yes if yy then no. Im not sure what xx could be in order to make it ask this.

Quote:

Originally Posted by XodoX

Thank you:)

It says thoug "want to try again?"

should I just make it reset ? If I just use printf to say yes or no it might be to simple. I dont know. Also dont know how I would do it though. Couldn't find a good C code database online.

Edit:

I gues I could do it with if. If xx then yes if yy then no. Im not sure what xx could be in order to make it ask this.

Usually a user query is done something like this:

Code:

do { print Enter your Sales Price scanf(" %f", &salesPrice); print Enter your Cash Paid etc. print Buy Another Item (y/n) ? scanf(" %c", &another); }while(another != 'n' || another != 'N')

This was what you were trying to show me before, right? I thought you were showing me something with a newline because of the backslash in it - \n.

Is this something I would use outside the classroom? Absolutely not. Scanf is just not well suited for the real world, these days. For a homework assignment, it's OK.

If you have more questions, please post your code so I can see what you're writing about. It's hard to understand the details from a posted description of something you can't see.

Code:

Quarters = Remaining / 25; // How many quarters? 0 Remaining = Remaining % 25; // How much left? 9 Dimes = Remaining / 10; // How many dimes? 0 Remaining = Remaining % 10; // How much left? 9 Nickels = Remaining / 5; // How many nickels? 1 Pennies = Remaining % 5; // How many pennies? 4

After this part there gotta be something that asks "Want to do this(try this) again?"Meaning want to calculate something again or not. Thats what I meant.
So basically this part would work .

Code:

print Buy Another Item (y/n) ? scanf(" %c", &another); }while(another != 'n' || another != 'n')

Which I would just add to the end.

Right. The do loop I posted above, needs to wrap around your posted code.

do

your code all goes here

want to buy another item?

while, etc

Nope. The do needs to go near the top of your logic, and the question "want to buy another item?" needs to go after all the other logic has finished. Then comes the while line.

One of those small n's needs to be a capital N - my mistake.

oh well this gets confusing now. But I am trying to understand.

Code:

do { SPDollars = (int)SalesPrice; // SPDollars = 5 SPCents = (int)(SalesPrice * 100.00) % 100; // 5.66 * 100.00 = 566.00; 566 % 100 = 66 cents CPDollars = (int)CashPaid; CPCents = (int)(CashPaid * 100.00) % 100; if (Remaining < 0) { // We had to borrow a dollar Dollars--; Remaining += 100; }; printf ( "Buy Another Item (y/n)" ) ? scanf(" %c", &another); }while(another != 'n' || another != 'N');

Thats how I understand it.

Unless you meant to put the code in yours and put it to the very end of the entire code.

Quote:

Originally Posted by XodoX

oh well this gets confusing now. But I am trying to understand.

Code:

do { SPDollars = (int)SalesPrice; // SPDollars = 5 SPCents = (int)(SalesPrice * 100.00) % 100; // 5.66 * 100.00 = 566.00; 566 % 100 = 66 cents CPDollars = (int)CashPaid; CPCents = (int)(CashPaid * 100.00) % 100; if (Remaining < 0) { // We had to borrow a dollar Dollars--; Remaining += 100; }; printf ( "Buy Another Item (y/n)" ) ? scanf(" %c", &another); }while(another != 'n' || another != 'N'); //replace || with &&

Thats how I understand it.

Unless you meant to put the code in yours and put it to the very end of the entire code.

Another correction: || should be replaced with && in the while loop.

The code to handle each purchase and return of change (if any), needs to go inside the do loop.

I don't know what you mean by "put the code in yours" ?? When you bring the code blocks together into the program, I believe you'll see where they need to go.

If not, post the code, and describe the problem *in detail*. You have an ability to leave me floating on a sea of question marks, from some of your posts.

Quote:

Originally Posted by Adak

Another correction: || should be replaced with && in the while loop.

The code to handle each purchase and return of change (if any), needs to go inside the do loop.

I don't know what you mean by "put the code in yours" ?? When you bring the code blocks together into the program, I believe you'll see where they need to go.

If not, post the code, and describe the problem *in detail*. You have an ability to leave me floating on a sea of question marks, from some of your posts.

No, if this is correct than everything is fine. The code who needs to be handled is in there now.
It looks like this now. So the single purchase is in it.

Code:

#include <iostream.h> main() { double SalesPrice = 0.0; double CashPaid = 0.0; int SPDollars; // Sales price dollars int SPCents; // Sales price cents int CPDollars; // Cash paid dollars int CPCents; // Cash paid cents int Dollars; // Number of $1 bills int Quarters; // Number of Quarters int Dimes; // Number of Dimes int Nickels; // Number of Nickels int Pennies; // Number of Pennies int Remaining; do { SPDollars = (int)SalesPrice; // SPDollars = 5 SPCents = (int)(SalesPrice * 100.00) % 100; // 5.66 * 100.00 = 566.00; 566 % 100 = 66 cents CPDollars = (int)CashPaid; CPCents = (int)(CashPaid * 100.00) % 100; if (Remaining < 0) { // We had to borrow a dollar Dollars--; Remaining += 100; }; printf ( "Buy Another Item (y/n)" ) ? scanf(" %c", &another); }while(another != 'n' && another != 'N') Quarters = Remaining / 25; // How many quarters? 0 Remaining = Remaining % 25; // How much left? 9 Dimes = Remaining / 10; // How many dimes? 0 Remaining = Remaining % 10; // How much left? 9 Nickels = Remaining / 5; // How many nickels? 1 Pennies = Remaining % 5; // How many pennies? 4 return 0; }

If you dont mean it like this I assume you mean the ENTIRE code in the loop instead of just my part but I hope like THIS is ok.

Like that is NOT OK, sorry.

For each purchase, you will have to have the code INSIDE the do loop, that gives back change, of course.

THEN you ask them if they'd like to make another purchase. :)

So drop this line of code, down below where you return the change (or print the amount of
change to be returned): (and you will print that up properly won't you?)

Code:

}while(another != 'n' && another != 'N')

Quote:

Originally Posted by Adak

Like that is NOT OK, sorry.

For each purchase, you will have to have the code INSIDE the do loop, that gives back change, of course.

THEN you ask them if they'd like to make another purchase. :)

Sorry but I don;t understand that at all. And I thought I finally got it:(
Now I understand it like the ENTIRE ode has to be in the loop.
I give up....:(

Nothing has to be returned. Just has to ask the question if the user wants to run the program again or not.

Quote:

Originally Posted by XodoX

Sorry but I don;t understand that at all. And I thought I finally got it:(
Now I understand it like the ENTIRE ode has to be in the loop.
I give up....:(

Nothing has to be returned. Just has to ask the question if the user wants to run the program again or not.

Not that bad:

Code:

#include <iostream.h> main() { double SalesPrice = 0.0; double CashPaid = 0.0; int SPDollars; // Sales price dollars int SPCents; // Sales price cents int CPDollars; // Cash paid dollars int CPCents; // Cash paid cents int Dollars; // Number of $1 bills int Quarters; // Number of Quarters int Dimes; // Number of Dimes int Nickels; // Number of Nickels int Pennies; // Number of Pennies int Remaining; do { SPDollars = (int)SalesPrice; // SPDollars = 5 SPCents = (int)(SalesPrice * 100.00) % 100; // 5.66 * 100.00 = 566.00; 566 % 100 = 66 cents CPDollars = (int)CashPaid; CPCents = (int)(CashPaid * 100.00) % 100; if (Remaining < 0) { // We had to borrow a dollar Dollars--; Remaining += 100; }; printf ( "Buy Another Item (y/n)" ) ? scanf(" %c", &another); //}while(another != 'n' && another != 'N') Quarters = Remaining / 25; // How many quarters? 0 Remaining = Remaining % 25; // How much left? 9 Dimes = Remaining / 10; // How many dimes? 0 Remaining = Remaining % 10; // How much left? 9 Nickels = Remaining / 5; // How many nickels? 1 Pennies = Remaining % 5; // How many pennies? 4 }while(another != 'n' && another != 'N') //from 7 lines up, to here, that's all. :) return 0; }

Put some good print statements in there to show what the Dollars, Quarter...Pennies, of change are, and aren't you all done?

Code:

#include <string.h> #include <stdlib.h> typedef struct { int dollars; int nickel; int pennies; }CHANGE; // Calculate Change CHANGE convertChange(char s_money[], float f_money) { char dot = '.'; CHANGE change; char *m_ptr, pennies[5]; // Get the dollar part change.dollars = f_money; // Find the decimal point m_ptr = strchr(s_money, dot); // And get the pennies part strcpy(pennies, ++m_ptr); change.pennies = atoi ( pennies ); // Not sure what is a nickle return change; }

would something like this work? Use Rands so not sure what nickle is....

Yes, that way was mentioned back on the first page of this thread iirc, and can work just fine (although I didn't run your code).

Clearly, you'd need to add Quarters - 25 pennies, dimes - 10 pennies, and nickels are worth 5 pennies, to your change struc.

This is a problem that can be solved a lot of different ways.