将浮点数转换为双精度整数

Question

I have to make a project that consists of calculating the amount of coins you need to form a number. You have quartes, dimes, nickels and pennies and you need to form a number with the least number of coins possible. The number you need to form is prompted to the user.

This is very simple in Python(I implement a function to do something quick):

def calculate_change(change):
    count = 0
    change*=100
    change = int(change)
    if change // 25 != 0:
        count = change // 25
        change = change % 25
    if change // 10 != 0:
        count += change // 10
        change = change % 10
    if change // 5 != 0:
        count += change // 5
        change = change % 5
    if change // 1 != 0:
        count += change // 1

    print count

calculate_change(73)

Now, to implement this in C, I'm having some troubles. I don't quite understand how to change the float the user prompted. I need to convert it first to double, multiply by 100 and then convert to an int but I'm having troubles with it. This is what I wrote for that:

int main(void)
{
    float n;
    do
    {
        printf("Change you need: ");
        n = GetFloat();
    }
    while (n < 0);

    n = floorf(n * 100 + 0.5) / 100;

    int change = (int) n;
    change = change * 100;
    int count = 0; 

    if (change / 25 != 0)
    {
        count = change / 25;
        change =  change % 25;
    }
    if (change / 10 != 0)
    {
        count = count + change / 10;
        change = change % 10;
    }
    if (change / 5 != 0)
    {   
        count = count + change / 5;
        change = change % 5;
    }
    if (change / 1 != 0)
    {
        count = count + change / 1;
        change = change % 1;
    }

    printf("%d\n", count);

}

Is there something wrong with the conversion from floats to ints in C ? The program returns incorrect results (always 0) when user prompts a number between 0 and 1.

Thanks !

Answer 1

Problem

When n = 0.5 ,

floorf(n * 100 + 0.5) / 100 = floorf(50.5)/100 = 50/100 = 0.5

Hence, the statement

n = floorf(n * 100 + 0.5) / 100;

does not change the value of n . Now, when you execute:

int change = (int) n;

change is set to 0 . That explains why you get 0 for most values of n between 0.0-1.0 .

Solution

I would replace the following lines:

n = floorf(n * 100 + 0.5) / 100;

int change = (int) n;
change = change * 100;

with

n = floorf(n * 100 + 0.5);  // Just to make sure rounding occurs properly.
                            // Thanks to Mark Ransom.
int change = (int) n;

Answer 2

You are right that you need to multiply by 100 and then convert to an int, and that's what you're doing in the python code, but in your C code you are doing it backwards - first converting to an int and then multiplying by 100:

 int change = (int) n;
 change = change * 100;

That (int) cast truncates values to the next lowest integer, so values of n between 0 and 1 will always get set to 0.

Moving the multiplication by 100 ensures that you convert from dollars (eg 0.56) to a floating point representation of cents (eg 56.0) before converting to an integer of cents eg (56):

int change = (int) (n * 100);