Programming Forums > Application Development > Software Design and Algorithms

peace_of_mind · Aug 13th, 2007, 10:46 AM

An assignment I'm working on has presented the following problem for me:
I have to calculate the change owed to a customer. No problem:

change = amountPaid - amountOwed

Here's where I'm having trouble. Not only do I have to display a message stating the change owed, but I have to display how many dollars, quarters, dimes, nickels, and pennies are owed. The logic for these calculations is where I'm getting bogged down. The way I initially approached it was to try:

dollars = change / 1
quarters = (change - dollars) / 0.25
dimes = (change - dollars - quarters) / 0.10
etc.

The problem with this is, obviously, the decimal remainder when the division occurs. I need a way to divide by each value, and then truncate the decimal remainder, leaving just the whole number for each calculation.
This project is in VB.NET. Any thoughts/suggestions would be appreciated.

peace_of_mind · Aug 13th, 2007, 11:55 AM

Okay I think I have figured it out. If I take the dollar amount and multiply it by 100 I get rid of the decimal. And as long as I multiply the other side of the equation by 100, I can use integer division to solve the problem. I think. When I get home to my source code I'll give this a try. In the meantime if anyone has any other approaches I'd love to hear them.

So the new code would look like:

(Toggle Plain Text)

Dim owedAmount, paidAmount, change, temp As Single
Dim dollars, quarters, dimes, nickels, pennies as integer

owedAmount = Val(Me.uiOwedAmountTextBox.Text)
paidAmount = Val(Me.uiPaidAmountTextBox.Text)

change = paidAmount - owedAmount
temp = change
dollars = (temp * 100) \ 100
temp = temp - dollars
quarters = temp * 100 \ 25
temp = temp - quarters * .25
etc.

DaWei · Aug 13th, 2007, 1:37 PM

This is VB, POM, but you can see the flow:

(Toggle Plain Text)

#include <iostream>

using std::cout;
using std::endl;

int main ()
{
    double owedAmount = 5.07;
    double paidAmount = 10.00;
    double change;

    int dollars, quarters, dimes, nickels;

    change = paidAmount - owedAmount;
    cout << "Owed: " << owedAmount << ", Paid: " << paidAmount << endl;
    cout << "Change: " << change << endl;

    dollars = change / 1;
    change = change - dollars;
    quarters = change / .25;
    change = change - quarters * .25;
    dimes = change / .10;
    change = change - dimes * .10;
    nickels = change / .05;
    change = change - nickels * .05;
    change = change / .01;

    cout << "Dollars: " << dollars << endl;
    cout << "Quarters: " << quarters << endl;
    cout << "Dimes: " << dimes << endl;
    cout << "Nickels: " << nickels << endl;
    cout << "Pennies: " << change << endl;

    return 0;
}

Quote:

Originally Posted by Output

Owed: 5.07, Paid: 10
Change: 4.93
Dollars: 4
Quarters: 3
Dimes: 1
Nickels: 1
Pennies: 3

peace_of_mind · Aug 13th, 2007, 2:02 PM

Thanks, DaWei. That's interesting. I'll have to go back and look over my original code before I try my idea from post #2. I used the same equations that you did, but I believe my variables were declared differently. Guess I gots some tinkering to do...

DaWei · Aug 13th, 2007, 5:18 PM

I would have mentioned fmod, but I don't know if VB has a similar function. The germane thing is that a float, when stuffed into an int, will lose its decimal places. You need to be aware that a floating point value cannot possibly represent all the values that lie in its range. That's because a value comprising a given number of bits can only represent 2^nbits distinct things. When only dealing with two decimal places, though, you should be reasonably safe.

peace_of_mind · Aug 13th, 2007, 8:59 PM

Thanks again. Once again I've learned something today.

For anyone interested here's how my code ended up. I'm rather pleased with it and I believe it's the approach the instructor was looking for.

(Toggle Plain Text)

        Dim owedAmount, paidAmount, change, temp As Decimal
        Dim dollars, quarters, dimes, nickels, pennies As Integer

        owedAmount = Me.uiOwedAmountTextBox.Text
        paidAmount = Me.uiPaidAmountTextBox.Text

        change = paidAmount - owedAmount
        temp = change
        dollars = (temp * 100) \ 100
        temp = temp - dollars
        quarters = (temp * 100) \ 25
        temp = temp - quarters * 0.25
        dimes = (temp * 100) \ 10
        temp = temp - dimes * 0.1
        nickels = (temp * 100) \ 5
        temp = temp - nickels * 0.05
        pennies = (temp * 100) \ 1
        temp = temp - pennies * 0.01

Seif · Aug 14th, 2007, 3:48 PM

glad you are pleased, must of taken all day to take out them there semi colons

In VB.Net you can set the precision of the Decimal data type. Also if you just wanted to display the number you can do FormatNumber() function. There is also ParseNumber() if you require the number from the previous string at any point.

Not as efficient as doing it Daweis way by any means but just an alternative for your consideration.

Aug 13th, 2007, 10:46 AM	#1
peace_of_mind Professional Programmer Join Date: Sep 2004 Location: Hell if I know most of the time Posts: 439 Rep Power: 5	Calculation problems with my VB.NET solution An assignment I'm working on has presented the following problem for me: I have to calculate the change owed to a customer. No problem: change = amountPaid - amountOwed Here's where I'm having trouble. Not only do I have to display a message stating the change owed, but I have to display how many dollars, quarters, dimes, nickels, and pennies are owed. The logic for these calculations is where I'm getting bogged down. The way I initially approached it was to try: dollars = change / 1 quarters = (change - dollars) / 0.25 dimes = (change - dollars - quarters) / 0.10 etc. The problem with this is, obviously, the decimal remainder when the division occurs. I need a way to divide by each value, and then truncate the decimal remainder, leaving just the whole number for each calculation. This project is in VB.NET. Any thoughts/suggestions would be appreciated. __________________ Amateurs built the ark Professionals built the Titanic

Aug 13th, 2007, 2:02 PM	#4
peace_of_mind Professional Programmer Join Date: Sep 2004 Location: Hell if I know most of the time Posts: 439 Rep Power: 5	Thanks, DaWei. That's interesting. I'll have to go back and look over my original code before I try my idea from post #2. I used the same equations that you did, but I believe my variables were declared differently. Guess I gots some tinkering to do... __________________ Amateurs built the ark Professionals built the Titanic

Aug 13th, 2007, 5:18 PM	#5
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	I would have mentioned fmod, but I don't know if VB has a similar function. The germane thing is that a float, when stuffed into an int, will lose its decimal places. You need to be aware that a floating point value cannot possibly represent all the values that lie in its range. That's because a value comprising a given number of bits can only represent 2^nbits distinct things. When only dealing with two decimal places, though, you should be reasonably safe. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Aug 13th, 2007, 8:59 PM	#6
peace_of_mind Professional Programmer Join Date: Sep 2004 Location: Hell if I know most of the time Posts: 439 Rep Power: 5	Thanks again. Once again I've learned something today. For anyone interested here's how my code ended up. I'm rather pleased with it and I believe it's the approach the instructor was looking for. (Toggle Plain Text) Dim owedAmount, paidAmount, change, temp As Decimal Dim dollars, quarters, dimes, nickels, pennies As Integer owedAmount = Me.uiOwedAmountTextBox.Text paidAmount = Me.uiPaidAmountTextBox.Text change = paidAmount - owedAmount temp = change dollars = (temp * 100) \ 100 temp = temp - dollars quarters = (temp * 100) \ 25 temp = temp - quarters * 0.25 dimes = (temp * 100) \ 10 temp = temp - dimes * 0.1 nickels = (temp * 100) \ 5 temp = temp - nickels * 0.05 pennies = (temp * 100) \ 1 temp = temp - pennies * 0.01 Dim owedAmount, paidAmount, change, temp As Decimal Dim dollars, quarters, dimes, nickels, pennies As Integer owedAmount = Me.uiOwedAmountTextBox.Text paidAmount = Me.uiPaidAmountTextBox.Text change = paidAmount - owedAmount temp = change dollars = (temp * 100) \ 100 temp = temp - dollars quarters = (temp * 100) \ 25 temp = temp - quarters * 0.25 dimes = (temp * 100) \ 10 temp = temp - dimes * 0.1 nickels = (temp * 100) \ 5 temp = temp - nickels * 0.05 pennies = (temp * 100) \ 1 temp = temp - pennies * 0.01 __________________ Amateurs built the ark Professionals built the Titanic

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Aug 13th, 2007, 11:55 AM	#2
peace_of_mind Professional Programmer Join Date: Sep 2004 Location: Hell if I know most of the time Posts: 439 Rep Power: 5	Okay I think I have figured it out. If I take the dollar amount and multiply it by 100 I get rid of the decimal. And as long as I multiply the other side of the equation by 100, I can use integer division to solve the problem. I think. When I get home to my source code I'll give this a try. In the meantime if anyone has any other approaches I'd love to hear them. So the new code would look like: (Toggle Plain Text) Dim owedAmount, paidAmount, change, temp As Single Dim dollars, quarters, dimes, nickels, pennies as integer owedAmount = Val(Me.uiOwedAmountTextBox.Text) paidAmount = Val(Me.uiPaidAmountTextBox.Text) change = paidAmount - owedAmount temp = change dollars = (temp * 100) \ 100 temp = temp - dollars quarters = temp * 100 \ 25 temp = temp - quarters * .25 etc. Dim owedAmount, paidAmount, change, temp As Single Dim dollars, quarters, dimes, nickels, pennies as integer owedAmount = Val(Me.uiOwedAmountTextBox.Text) paidAmount = Val(Me.uiPaidAmountTextBox.Text) change = paidAmount - owedAmount temp = change dollars = (temp * 100) \ 100 temp = temp - dollars quarters = temp * 100 \ 25 temp = temp - quarters * .25 etc. __________________ Amateurs built the ark Professionals built the Titanic

Aug 14th, 2007, 3:48 PM	#7
Seif Hobbyist Programmer Join Date: Jan 2006 Location: UK Posts: 215 Rep Power: 3	glad you are pleased, must of taken all day to take out them there semi colons In VB.Net you can set the precision of the Decimal data type. Also if you just wanted to display the number you can do FormatNumber() function. There is also ParseNumber() if you require the number from the previous string at any point. Not as efficient as doing it Daweis way by any means but just an alternative for your consideration.

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Programming Forums > Application Development > Software Design and Algorithms

Calculation problems with my VB.NET solution