Moving decimal place to right in c

Question

I'm new to C and when I run the code below, the value that is put out is 12098 instead of 12099.

I'm aware that working with decimals always involves a degree of inaccuracy, but is there a way to accurately move the decimal point to the right two places every time?

#include <stdio.h>

int main(void)
{
    int i;
    float f = 120.99;

    i = f * 100;
    printf("%d", i);
}

Answer 1

Use the round function

float f = 120.99;
int i = round( f * 100.0 );

Be aware however, that a float typically only has 6 or 7 digits of precision, so there's a maximum value where this will work. The smallest float value that won't convert properly is the number 131072.01. If you multiply by 100 and round, the result will be 13107202.

You can extend the range of your numbers by using double values, but even a double has limited range. (A double has 16 or 17 digits of precision.) For example, the following code will print 10000000000000098

double d = 100000000000000.99;
uint64_t j = round( d * 100.0 );
printf( "%llu\n", j );

That's just an example, finding the smallest number is that exceeds the precision of a double is left as an exercise for the reader.

Answer 2

Use fixed-point arithmetic on integers:

#include <stdio.h>

#define abs(x) ((x)<0 ? -(x) : (x))

int main(void)
{
    int d = 12099;
    int i = d * 100;
    printf("%d.%02d\n", d/100, abs(d)%100);
    printf("%d.%02d\n", i/100, abs(i)%100);
}

Answer 3

Your problem is that float are represented internaly using IEEE-754. That is in base 2 and not in base 10. 0.25 will have an exact representation, but 0.1 has not, nor has 120.99.

What really happens is that due to floating point inacuracy, the ieee-754 float closest to the decimal value 120.99 multiplied by 100 is slightly below 12099, so it is truncated to 12098. You compiler should have warned you that you had a truncation from float to in (mine did).

The only foolproof way to get what you expect is to add 0.5 to the float before the truncation to int :

i = (f * 100) + 0.5

But beware floating point are inherently inaccurate when processing decimal values.

Edit :

Of course for negative numbers, it should be i = (f * 100) - 0.5 ...

Answer 4

If you'd like to continue operating on the number as a floating point number, then the answer is more or less no. There's various things you can do for small numbers, but as your numbers get larger, you'll have issues.

If you'd like to only print the number, then my recommendation would be to convert the number to a string, and then move the decimal point there. This can be slightly complicated depending on how you represent the number in the string (exponential and what not).

If you'd like this to work and you don't mind not using floating point, then I'd recommend researching any number of fixed decimal libraries.