Is there a way to convert 0.15 to 15 - c

Question

I want to convert a floating point number like 0.15 -> 15 or 0.4 -> 4
My idea is to
Multiply a number with amount of floating point places.
Floor it.
Cast it to (int).
Can't figure how to do the first step.
Clarification: I have to make a program that uses a greedy algorithm, User enters change_owed. This program would then check for the lowest number of coins possible to give-out.
Code:
#include <ctype.h>
#include <stdio.h>
int main(void)
{
int coins = 0;
float cash_owed = 0;
do
{
printf("Cash owed: ");
scanf("%f" , &cash_owed);
}
while(cash_owed < 0);
while(cash_owed > 0)
{
if(cash_owed >= 0.25)
{
cash_owed -= 0.25;
printf("Cash owed: %f\n", cash_owed);
coins++;
printf("Coins: %d\n", coins);
}
else if(cash_owed >= 0.10)
{
cash_owed -= 0.10;
printf("Cash owed: %f\n", cash_owed);
coins++;
printf("Coins: %d\n", coins);
}
else if(cash_owed >= 0.05)
{
cash_owed -= 0.05;
printf("Cash owed: %f\n", cash_owed);
coins++;
printf("Coins: %d\n", coins);
}
else
{
cash_owed -= 0.01;
printf("Cash owed: %f\n", cash_owed);
coins++;
printf("Coins: %d\n", coins);
}
}
printf("%d\n", coins);
return 0;
}
Not my best design. But it fails on many different examples given below, Acc to my understanding it's due to floating-point arithmetic imprecision.
Fails for: 0.15, 1.6 and 4.2. wherein it skips one or more if statement due to floating-point arithmetic imprecision.
So I devised a new algo which will convert these values to proper int. Like
0.15 -> 15
So I have to work with integers to eliminate that float point errors, If this makes any sense

Answer 1

to convert a floating point number like 0.15 -> 15 or 0.4 -> 4
Recognize 0.15 and 0.4 are not exactly encodable as a floating point number as they are not dyadic as typically used: e.g. binary32. Instead a nearby value is saved in cash.
Multiply the floating point number by 100. Since the product determination may round and the original number may only be close to a few digit decimal number, round the result. Using floor may take a value of xxx.9999... to xxx.0 rather than a more reasonable xxx.0 + 1.
#include <math.h>
long cash_in_cets = lround(cash * 100.0);
Proceed with rest of greedy algorithm using integer math and integer cents,
My crystal ball hints that the next version of C will specify decimal floating point numbers, which is a great solution for money.
Code can stay with only using FP types, yet test like (cash_owed >= 0.10) are too simplistic/wrong and need significant rework. Best to move to integer math.

Answer 2

You don't need to convert from floating point to integer to get the program to work.
From what I could see, the problem is that your program has no idea what to do once it gets to 0, probably because it never actually reaches absolute zero. One way to get around this is to add an additional else if statement that sets cash_owed to less than 0 whenever it gets less than 0.01:
else if(cash_owed < 0.01)
{
cash_owed = -1;
}
You also need to add a condition to the existing 0.01 else if statement:
else if(cash_owed >= 0.01)
{
cash_owed -= 0.01;
printf("Cash owed: %f\n", cash_owed);
coins++;
printf("Coins: %d\n", coins);
}
So your while loop end up looking like this:
while(cash_owed > 0)
{
if(cash_owed >= 0.25)
{
cash_owed -= 0.25;
printf("Cash owed: %f\n", cash_owed);
coins++;
printf("Coins: %d\n", coins);
}
else if(cash_owed >= 0.10)
{
cash_owed -= 0.10;
printf("Cash owed: %f\n", cash_owed);
coins++;
printf("Coins: %d\n", coins);
}
else if(cash_owed >= 0.05)
{
cash_owed -= 0.05;
printf("Cash owed: %f\n", cash_owed);
coins++;
printf("Coins: %d\n", coins);
}
else if(cash_owed >= 0.01)
{
cash_owed -= 0.01;
printf("Cash owed: %f\n", cash_owed);
coins++;
printf("Coins: %d\n", coins);
}
else if(cash_owed < 0.01)
{
cash_owed = -1;
}
}
This should work as long as your user enters valid amounts (so you might want to add a check to make sure they don't add more than 2 decimals).