Velocity Reviews > Beginner question: Precision of floating point arithmetic...

# Beginner question: Precision of floating point arithmetic...

Shawn
Guest
Posts: n/a

 07-14-2004
Hello all,
I apologize as I am sure this has probably been dealth with before...
but I am doing an exercise from "Practical C Programming" and I have
been unable to get it to work perfectly due to problems with floating
point arithmetic and I am looking for a way to solve it. See the code
below...
Given a certain amount of change (below \$1.00) the program will tell
you how many of each coin you will need to get that amount. The
program is "working" in that the logic appears correct and it DOES
work for some numbers, but for others, it is not. The problem appears
to be that 0.01 as I see it, is not being represented in memory.
I have done some searching online and I am sure this is a common
problem but I just can't seem to find the workaround...

Thank you for all your help...

Shawn

#include <stdio.h>

int main()
{
char input[100];
float total_input, running_total;
int quarters, nickels, dimes, pennies;

/* Zero out all the variables */
quarters = nickels = dimes = pennies = 0;
total_input = running_total = 0;

/* Get the necessary amount of change */
printf("Enter the total amount of change, less than \$1.00: ");
fgets(input, sizeof(input), stdin);
sscanf(input, "%f", &total_input);

/* Loop until we get a sane amount */
while (total_input >= 1.00 || total_input <= 0.00) {
printf("Total is not between \$0.00 and \$1.00.\n");
printf("Enter the total amount of change, less than
\$1.00: ");
fgets(input, sizeof(input), stdin);
sscanf(input, "%f", &total_input);
}

/* Store in another variable so we can use it */
running_total = total_input;

while (running_total >= 0.25) {
++quarters;
running_total -= 0.25;
}
while (running_total >= 0.10) {
++dimes;
running_total -= 0.10;
}
while (running_total >= 0.05) {
++nickels;
running_total -= 0.05;
}
while (running_total >= 0.01) {
++pennies;
running_total -= 0.01;
}

printf("In order to get \$%.2f in change, you will need:\n",
total_input);
printf("%d quarters,\n", quarters);
printf("%d dimes,\n", dimes);
printf("%d nickels,\n", nickels);
printf("%d pennies\n", pennies);

return(0);
}

Eric Sosman
Guest
Posts: n/a

 07-14-2004
Shawn wrote:
> Hello all,
> I apologize as I am sure this has probably been dealth with before...
> but I am doing an exercise from "Practical C Programming" and I have
> been unable to get it to work perfectly due to problems with floating
> point arithmetic and I am looking for a way to solve it. See the code
> below...
> Given a certain amount of change (below \$1.00) the program will tell
> you how many of each coin you will need to get that amount. The
> program is "working" in that the logic appears correct and it DOES
> work for some numbers, but for others, it is not. The problem appears
> to be that 0.01 as I see it, is not being represented in memory.
> I have done some searching online and I am sure this is a common
> problem but I just can't seem to find the workaround...

First, the question has, as you suspect, arisen before.
In fact, it arises frequently, and therefore has a place in
the comp.lang.c Frequently Asked Questions (FAQ) list

http://www.eskimo.com/~scs/C-faq/top.html

Once you've digested that, you'll have realized that
floating-point arithmetic is trickier than it first appears.
Floating-point is very good at dealing with proportions and
ratios and logarithms and the like, but is not well-suited
to counting problems -- problems involving discrete "things,"
if you like. So what's the work-around? Well, can you think
of a way to restate your original problem in terms of discrete
indivisible units instead of fractions of something-or-other?
in change?

--
http://www.velocityreviews.com/forums/(E-Mail Removed)

Malcolm
Guest
Posts: n/a

 07-14-2004

"Shawn" <(E-Mail Removed)> wrote in message
>
> #include <stdio.h>
>
> int main()
> {
> char input[100];
> float total_input, running_total;
> int quarters, nickels, dimes, pennies;
>
> /* Zero out all the variables */
> quarters = nickels = dimes = pennies = 0;
> total_input = running_total = 0;
>
> /* Get the necessary amount of change */
> printf("Enter the total amount of change, less than \$1.00: ");
> fgets(input, sizeof(input), stdin);
> sscanf(input, "%f", &total_input);
>

You want to check your return from sscanf().
>
> /* Loop until we get a sane amount */
> while (total_input >= 1.00 || total_input <= 0.00) {
> printf("Total is not between \$0.00 and \$1.00.\n");
> printf("Enter the total amount of change, less than
> \$1.00: ");
> fgets(input, sizeof(input), stdin);
> sscanf(input, "%f", &total_input);
> }
>
> /* Store in another variable so we can use it */
> running_total = total_input;
>
> while (running_total >= 0.25) {
> ++quarters;
> running_total -= 0.25;
> }
>

Here's your problem. Floating point representation is not exact, so if the
user enters 0.50 theres no guarantee that this won't be represented
internally as 0.4999, which will mess up your logic.
In fact all floating point units use a binary base, so 0.50 and 0.25 can be
represented exactly, but 0.1 cannot be. So the problem isn't here but in the
similar block of code for the dimes, below.
>
> while (running_total >= 0.10) {
> ++dimes;
> running_total -= 0.10;
> }
> while (running_total >= 0.05) {
> ++nickels;
> running_total -= 0.05;
> }
> while (running_total >= 0.01) {
> ++pennies;
> running_total -= 0.01;
> }
>
> printf("In order to get \$%.2f in change, you will need:\n",
> total_input);
> printf("%d quarters,\n", quarters);
> printf("%d dimes,\n", dimes);
> printf("%d nickels,\n", nickels);
> printf("%d pennies\n", pennies);
>
> return(0);
> }

>

The solution is to convert your float value, input, into an integer number
of pennies (I thought you Americans used cents, but that's by the by).

To do this, multiply the float value by 100, then round by adding 0.5 and
calling floor(). Just multiplying by 100 risks the inaccuracy problem
because a value of x.9999 will be rounded down.

Gordon Burditt
Guest
Posts: n/a

 07-14-2004
>I apologize as I am sure this has probably been dealth with before...
>but I am doing an exercise from "Practical C Programming" and I have
>been unable to get it to work perfectly due to problems with floating
>point arithmetic and I am looking for a way to solve it. See the code
>below...
>Given a certain amount of change (below \$1.00) the program will tell
>you how many of each coin you will need to get that amount. The

Use integer quantities of cents. You may use a floating-point variable
to store this if you want.

>program is "working" in that the logic appears correct and it DOES
>work for some numbers, but for others, it is not. The problem appears
>to be that 0.01 as I see it, is not being represented in memory.

There is no exact representation for 0.01 in binary floating point, nor
is there for most non-integer decimal numbers. (Think about representing
1/3 exactly in decimal). There *IS* a representation, just not an
exact one.

>I have done some searching online and I am sure this is a common
>problem but I just can't seem to find the workaround...

It's not exact. Learn to live with it. The world is also an analog
world with all sorts of floating-point measurements, and no measurement
is exact. Comparing two floating-point numbers for exact equality
is hazardous. How close two floating-point numbers have to be
(either by ratio or by absolute difference) generally should be
APPLICATION-DEPENDENT, and NOT depend on details like the number
of significant digits being used in the implementation. Example:
the GPS coordinates of two telephone poles are considered identical
(and to refer to the same pole) if they are within 20 feet of each
other, because GPS cannot reliably measure distances more accurately
than that, and besides, phone companies don't place poles that close
to each other.

One approach for money is to use the integer value 1 to represent the smallest
amount of money of interest (which for some applications might be a tenth
or thousandth of a cent in the USA). Of course, printing such a value
on checks needs to be done carefully.

Gordon L. Burditt

Keith Thompson
Guest
Posts: n/a

 07-15-2004
"Malcolm" <(E-Mail Removed)> writes:
[...]
> The solution is to convert your float value, input, into an integer number
> of pennies (I thought you Americans used cents, but that's by the by).

<OT>
A "cent" is a unit of currency equal to \$0.01 (one hundredth of a
dollar). A "penny" is an informal name for the one-cent coin; the
term is also sometimes used as a synonym for "cent". Similarly,
"nickel" is the informal name for a five-cent coin. A "dime" is
actually a unit of currency equal to \$0.10, or 10 cents, but these
days the term is used almost exclusively to refer to the 10-cent coin.
</OT>

--
Keith Thompson (The_Other_Keith) (E-Mail Removed) <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

Richard Bos
Guest
Posts: n/a

 07-15-2004
Eric Sosman <(E-Mail Removed)> wrote:

> Hint: What should your program do if someone asks for \$0.14159
> in change?

Sell him some more pie?

Richard

Joona I Palaste
Guest
Posts: n/a

 07-15-2004
Keith Thompson <(E-Mail Removed)> scribbled the following:
> "Malcolm" <(E-Mail Removed)> writes:
> [...]
>> The solution is to convert your float value, input, into an integer number
>> of pennies (I thought you Americans used cents, but that's by the by).

> <OT>
> A "cent" is a unit of currency equal to \$0.01 (one hundredth of a
> dollar). A "penny" is an informal name for the one-cent coin; the
> term is also sometimes used as a synonym for "cent". Similarly,
> "nickel" is the informal name for a five-cent coin. A "dime" is
> actually a unit of currency equal to \$0.10, or 10 cents, but these
> days the term is used almost exclusively to refer to the 10-cent coin.
> </OT>

Similarly, I've heard that the British use "pence" to refer a monetary
sum, but "pennies" to refer to a collection of coins.

--
/-- Joona Palaste ((E-Mail Removed)) ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"We're women. We've got double standards to live up to."
- Ally McBeal

Alex Fraser
Guest
Posts: n/a

 07-15-2004
"Joona I Palaste" <(E-Mail Removed)> wrote in message
news:cd5q6i\$ajp\$(E-Mail Removed)...
[snip]
> Similarly, I've heard that the British use "pence" to refer a monetary
> sum, but "pennies" to refer to a collection of coins.

Yes. The coins are labeled "ONE PENNY". Paradoxically, older ones say "NEW
PENNY" .

Alex

Lew Pitcher
Guest
Posts: n/a

 07-15-2004
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Alex Fraser wrote:
> "Joona I Palaste" <(E-Mail Removed)> wrote in message
> news:cd5q6i\$ajp\$(E-Mail Removed)...
> [snip]
>
>>Similarly, I've heard that the British use "pence" to refer a monetary
>>sum, but "pennies" to refer to a collection of coins.

>
>
> Yes. The coins are labeled "ONE PENNY". Paradoxically, older ones say "NEW
> PENNY" .

And, what are the ones issued before 1971 called?

(hint: http://en.wikipedia.org/wiki/Pound_Sterling)

- --
Lew Pitcher
IT Consultant, Enterprise Application Architecture,
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed are my own, not my employers')
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Alex Fraser
Guest
Posts: n/a

 07-15-2004
"Lew Pitcher" <(E-Mail Removed)> wrote in message
news:8uAJc.26706\$(E-Mail Removed) ...
> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>
> Alex Fraser wrote:
> > "Joona I Palaste" <(E-Mail Removed)> wrote in message
> > news:cd5q6i\$ajp\$(E-Mail Removed)...
> > [snip]
> >
> >>Similarly, I've heard that the British use "pence" to refer a monetary
> >>sum, but "pennies" to refer to a collection of coins.

> >
> >
> > Yes. The coins are labeled "ONE PENNY". Paradoxically, older ones say
> > "NEW PENNY" .

>
> And, what are the ones issued before 1971 called?

Annoyingly large?

Alex