Change Maker (problems with %)

OK, so for school, I have to make a program that illustrates the change a person should get based on a few monetary inputs. I'm crafting one that does so in the following steps:

1. User inputs monetary value (e.g. 17.34 for $17.34) describing the original bill.

2. User inputs monetary value describing how much cash they paid.

3a. If the amount paid is less than the bill, it tells you that you still owe the difference (which is produced).

3b. If the amount paid is more than the bill, it illustrates the most efficient change you should get, in dollars, quarters, dimes, nickels, and pennies.

3b is where I'm stuck. I know that inputting, for example, (25 % 3) would return the remainder of 25 / 3, which is 1. However, I want to divide, say, 25 by 3, return the 8, and keep the one to work with in the next step.

This is confusing me. D:

To get the eight, you would simply use integer math, ie 25 / 3 produces 8. 25 % 3 produces the one. Simply save them or whatever and continue processing.
If you cannot grasp how to do that, then you need to write down the logical steps needed to perform that. Imagine someone owing you something, for example, and think logically how such transactions would be done.
Btw, don't use goto.

Quote:

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Originally Posted by Elysia

Btw, don't use goto.

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I SAID NOT TO TERRORIZE ME! :'(

So using the variables as integers in division just rounds them to the nearest whole number?

More precisely, it truncates the decimals. Eg 11 / 6 = 1.

Oh, so it always rounds down?
OK I'll give it a shot.

No, it discards the decimals. Chops them off.
Not quite the same thing, but I suppose you could say that.

I see what you mean.

So I say:

Code:

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cout << setprecision(0) << fixed << ((paid - bill) % 1);

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?
I assume that setprecision(0) would chop off the decimals as well. Kind of easier to manage how many decimal places for each statement individually, right?

Would I use % or / 1?

The for the next one, I'm not sure how to call in the leftovers from the last function.

There is no need to use % 1 or / 1 at all. They will not do anything.

Quote:

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The for the next one, I'm not sure how to call in the leftovers from the last function.

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You will need to formulate proper logic for it first. Eg, how would you do it, in real life?

I'm not trying to be difficult, but I'm really new to programming so I probably don't know half the objects or statements you're trying to get me to think of. I see where you're coming from, not spoon-feeding everyone answers and all, and I respect that, but I could use more straightforward answers to learn much better.

Not getting the objects or statements, or being really new isn't much of an excuse though. People who say things like that eventually get used to giving up at the slightest hardship and not doing anything. Not to mention this problem is entirely math based: if you paid $20 for a 17.34 bill then you could get $2, 2 quarters, 1 dime, 1 nickel, and 1 penny back. What did I really divide to get that, and why would you want to go on to a more difficult problem than this one, without figuring it out?


You ask how you want to save the result of 25 / 3 or 25 % 3. What are variables for?

Quote:

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Originally Posted by muffinman8641

I'm not trying to be difficult, but I'm really new to programming so I probably don't know half the objects or statements you're trying to get me to think of. I see where you're coming from, not spoon-feeding everyone answers and all, and I respect that, but I could use more straightforward answers to learn much better.

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If you really don't understand what they are telling you, try to understand that.... You can't live with the knowledge you gain in your class rooms.... Try to get your own.... Internet is the best source of getting knowledge......

Good luck..

% 1 gives the remainder of dividing something by one, which of course is always zero.

Quote:

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Originally Posted by muffinman8641

I'm not trying to be difficult, but I'm really new to programming so I probably don't know half the objects or statements you're trying to get me to think of. I see where you're coming from, not spoon-feeding everyone answers and all, and I respect that, but I could use more straightforward answers to learn much better.

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All I'm saying is this:
Before getting into statements and objects and whatnot, you should think of how the program should behave. Ie, how do you WANT it to behave? The logical part.
Then we can get into translating the logical part into statements and objects.

OK, sorry if I came off as an ass there.

I need to take the difference between the paid amount and the bill, then break it down into dollars and coins.

I need to say "How many whole dollars fit into this?"
Then subtract the dollars and say "How many quarters are in the remains?"
etc.

[Added note- I tried the suggestion of

Code:

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int x, y;
double a, b;

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for splitting up variable types, and it seemed to work ok. I'm just stuck on how to get a whole number with the decimal chopped off, because bill and paid are both originally double variables, since there's change involved with both. I tried making a new int variable called intdollars = (paid - bill), but I got a warning about that and it returned -2 billion dollars when I tested it.]

The reason will be that paid and bill will be double datatype variables and double datatype occupies 8 bytes in the memory (C++), while integer just occupies 4 bytes (C++), so it truncates or process something differently while performing that operation...

You must declare dollars variable as double datatype and then you can format the decimal point using flags or setprecision....
Or

Code:

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double num = 23.345;
 int intpart = (int)num;

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Quote:

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Originally Posted by Mr.777

You must declare dollars variable as double datatype and then you can format the decimal point using flags or setprecision....

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Good idea! I made my code this:

Code:

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double bill, paid, dollars = (paid - bill);

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and also this:

Code:

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cout << "The most efficient way to get your change would be to get:" << endl;
            cout << setprecision(0) << fixed <<  dollars << " dollars, " << endl;

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Problem is, when I'm testing the program, it says "The most efficient way to get your change would be to get:", then nothing. Then it says the program stopped working. Basically, it crashes.

Because you've just declared the variables bill and paid but didn't assign any value to them... So, do;

Code:

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double bill, paid;
bill=some_Value;
paid=some_Other_Value;
double dollars=paid-bill;

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The user defines the value of the bill and paid in the program.

Then,

Code:

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double paid,bill;
cin>>paid>>bill;
double dollars=paid-bill;

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Try to think pleaseeeeeeeeeeeee...........

I figured that part out, just one last question and this program will (hopefully) be complete;
how do I display a math result (e.g. (bill - paid)), then save the answer for use in the next math computation? Return?

Oh God.......
Don't you know, how to show something over console (monitor)??? If you really don't, just read then.

Again, saving results typically means you remember something, and in the computer world, that means saving the result to a variable. How and when? That's for your logical institution to decide.

Is there a way to display ONLY the decimals of a number? I'm just asking odd questions like this here and there trying different ideas; I googled it with no results.

For example, if the number was 5.68, how could I get the program to display ONLY the 0.68? Don't say subtract 5 either, it will be different almost every time.

modf is one possibility.

OK, here's my code:

Code:

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#include <iostream>
#include <iomanip>
#include <time.h>

using namespace std;

int main() {
   

    double bill, paid;
    int list;
   
    cout << "Welcome to my change-maker program! " << endl;
    cout << "To begin, press ENTER. " << endl;
    getchar();
   
    restart:
   
    cout << "\nEnter how much the bill is. " << endl;
    cout << "Do not use a $ sign. Just enter the decimal number. " << endl;
    cin >> bill;
    cout << "Now enter how much cash you paid. " << endl;
    cin >> paid;
    if (paid < bill) {
            cout << "Uh-oh! You need to pay more cash. You " << endl;
            cout << "haven't paid the whole bill yet. " << endl;
            cout << "You still owe " << setprecision(2) << fixed << (bill - paid) << ". " << endl;
            goto restart;
            }
   
    if (bill <= paid) {
            cout << "The most efficient way to get your change would be to get:" << endl;
            cout << setprecision(0) << fixed << (paid - bill) << " dollars, " << endl;
            cout << "----------------------------------------------" << endl;
            cout << "TOTAL: " << (paid - bill) << endl;

}
list_reset:
cout << "I would like to... " << endl;
cout << "[1] Restart the program " << endl;
cout << "[2] Quit the program " << endl;
cout << "[3] Receive a compliment, then quit the program " << endl;

cin >> list;

if ((list < 1) || (list > 3)) {
            cout << "Please enter a number ranging from one to three. " << endl;
            goto list_reset;
            }
           
if (list == 1) {
goto restart;
}

if (list == 2) {
  cout << "Bye. " << endl;
}

if (list == 3) {
    cout << "\nI like your shirt. " << endl;
}
   
end_program:

system ("pause");
return (0);
}

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As you can see, there's a part that says:

Code:

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cout << "The most efficient way to get your change would be to get:" << endl;
cout << setprecision(0) << fixed << (paid - bill) << " dollars, " << endl;

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Below that is where I'd like to stick the rest with the quarters and change.

I tried modf, but I was confused and messed it up. I'll look more into it.

Other than the imminent criticisms about goto, what can I do to it to make it do what I want?

Calculate dollars it's possible to give back, then calculate whatever sub-dollars it's possible to give back, and so on.
Just like turning seconds into hours, minutes and seconds. Piece of cake.

I think modf will be the best option for you in this case.....
I am giving you an example of using modf...

Code:

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double integer_Number, decimal_Number;
double decimal_Number=modf(5.68,integer_Number);
// This will return 0.68 to the decimal_Number and 5 to the integer_Number.

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Try using this....

That should be

Code:

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double integer_Number, decimal_Number;
double decimal_Number=modf(5.68, &integer_Number);
// This will return 0.68 to the decimal_Number and 5 to the integer_Number.

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My bad :-)

Code:

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cout << "The most efficient way to get your change would be to get:" << endl;
            double integer_number;
            double decimal_number = modf((bill - paid), &integer_number);
            cout << setprecision(0) << fixed << (bill - paid) << " dollars, \n";
            cout << setprecision(0) << fixed << (decimal_number / 4) << " quarters, \n";
            cout << setprecision (0) << fixed << ((integer_number / 4) / 10) << " dimes, \n" ;
            cout << setprecision (0) << fixed << (((integer_number / 4) / 10) /5) << " nickels, and\n";
            cout << setprecision (0) << fixed << ((((integer_number / 4) / 10) /5) / 1) << " pennies.\n" ;
            cout << "----------------------------------------------" << endl;

---

It returns the dollars, but then 0 for all the coins. It also puts a hyphen (-) before each one, as if it's negative, but also the zeroes.
[Edit: noticed that I did bill - paid. Should be paid - bill]

Any corrections?

If something costs $50, and I pay $50.5, do you get .5 subdollars or 5 subdollars, or 5x subdollars, where x is 10 raised to a number of choice?
modf returns the decimals, so 50.5 becomes 50 and .5.

I don't know, why you are not trying to think.... Just play with the functions with different values, not just hit and miss......

Code:

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double paid=50.57;
double bill=50.0;
double integer_Number, double_Number;
double_Number=modf((paid-bill),&integer_Number);
cout<<"Your total bill was:"<<bill<<" and you've paid"<<paid<<". So the remaining amount is"<<double_Number<<endl;

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What's so difficult?????
What are the problems???
If there are, you should better try to think logically...... Just don't learn something new from anywhere and if you couldn't do, again go back... First learn good, read about it.... Then implement it with different examples (I mean play with that) and then if you still got problems, share them.....

Quote:

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Originally Posted by Mr.777

I don't know, why you are not trying to think.

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I actually am. I've been thinking about this for days. I don't know as much abut functions and the options I have as the mods and users here do, so that's why I started the thread.

It works! I have to stick a statement in there to see if it adds up and if not add a penny, because once in a while it's off. Here's the code, sans the ASCII bunny I put in the final version. :p My teacher has a sense of humor, so it's not like I'll get in trouble for it.

Code:

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#include <iostream>
#include <iomanip>
#include <time.h>
#include <stdio.h>
#include <math.h>

using namespace std;

int main() {
   

    double bill, paid;
    int list;
   
    cout << "CHANGE MAKER" << endl;
    cout << "To begin, press ENTER. " << endl;
    getchar();
   
    restart:
   
    cout << "\nEnter how much the bill is. " << endl;
    cout << "Do not use a $ sign. Just enter the decimal number. " << endl;
    cin >> bill;
    cout << "Now enter how much cash you paid. " << endl;
    cin >> paid;
    if (paid < bill) {
            cout << "Uh-oh! You need to pay more cash. You " << endl;
            cout << "haven't paid the whole bill yet. " << endl;
            cout << "You still owe " << setprecision(2) << fixed << (bill - paid) << ". " << endl;
            goto restart;
            }
   
    if (bill <= paid) {
            cout << "The most efficient way to get your change would be to get" << endl;
            int dollars = (paid - bill), quarters = (((paid - bill) - dollars) / 0.25);
            int dimes = ((((paid - bill) - dollars) - (quarters * 0.25)) / 0.1);
            int nickels = ((((paid - bill) - dollars) - (quarters * 0.25) - (dimes * 0.1)) / 0.05);
            int pennies = ((((paid - bill) - dollars) - (quarters * 0.25) - (dimes * 0.1) - (nickels * 0.05)) / 0.01);
            cout << dollars << " dollar(s), " << endl;
            cout << quarters << " quarter(s), " << endl;
            cout << dimes << " dime(s), " << endl;
            cout << nickels << " nickel(s), and " << endl;
            cout << pennies << " pennies(s). " << endl;
            cout << "----------------------------------------------" << endl;
            cout << "TOTAL CHANGE: " << (paid - bill) << endl;
            cout << "You may have noticed that the number of pennies is incorrect. " << endl;
            cout << "This sometimes happens. Just add one. " << endl;
}
list_reset:
cout << "\n\nI would like to... " << endl;
cout << "[1] Restart the program " << endl;
cout << "[2] Quit the program " << endl;
cout << "[3] Receive a compliment, then quit the program " << endl;
cout << "[4] See an ASCII picture of a deranged bunny " << endl;

cin >> list;

if ((list < 1) || (list > 3)) {
            cout << "Please enter a number ranging from one to three. " << endl;
            goto list_reset;
            }
           
if (list == 1) {
goto restart;
}

if (list == 2) {
  cout << "Bye. " << endl;
}

if (list == 3) {
    cout << "\nI like your shirt. " << endl;
}
   

end_program:

system ("pause");
return (0);
}

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Mr. 777, I'd appreciate it if you'd either help or stop talking, because I don't appreciate being talked down to because I don't understand this concept.

T'would be easier if you just took calculated the amount of dollars, then take the rest and calculate the nickels or whatever, and so on.
And get rid of the goto already.

Quote:

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Mr. 777, I'd appreciate it if you'd either help or stop talking, because I don't appreciate being talked down to because I don't understand this concept.

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First thing, all my advices were just to make you think logically.
Scond thing, do i care???

Quote:

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It works! I have to stick a statement in there to see if it adds up and if not add a penny, because once in a while it's off.

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Yeah, some numbers cannot be represented exactly in floating point. When accuracy is needed, as is the case with finance, you can always express amounts in terms of pennies.

long money = int(100 * amount + 0.5);

That .5 actually adds the penny for goofy amounts.

What you said you do instead is OK of course.

And what if paid is greater than bill?
I think you should handle that too....

Quote:

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Originally Posted by Mr.777

And what if paid is greater than bill?
I think you should handle that too....

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Paid being greater than the bill is the whole program. If you mean paid being LESS than the bill, there's a chunk in there for such events. Take a look and you'll find it.