How to get the exact fractional part from a floating point number as an integer?

Question

I was trying to extract the exact fractional part from a floating point number. I tried with this:

float f=254.73;

int integer = (int)f;
float fractional = f-integer;

printf ("The fractional part is: %f", fractional);

But the output is: 0.729996. For this reason when I was doing this:

float f=254.73;

int integer = (int)f;
float fractional = f-integer;
int fractional_part_in_integer = ((int)(f*100)%100);

printf ("The value is: %d", fractional_part_in_integer);

It gives me 72 as output. But, I want to extract exactly 73 from the given number 254.73. I already know how to use %.2f during printf() function to print upto two decimal numbers. But in my code I don't want to print the number right now. I have some calculations with that fractional part as integer form i.e. 73.

So, my problem is how could I extract the fractional part from 254.73 so that I can get exact 73 as integer to do more calculations?

Answer 1

How to get the exact fractional part from a floating point number as an integer?

trying to extract the exact fractional part from a floating point number.

Use modf() or modff()

double modf(double value, double *iptr);
float modff(float value, float *iptr);

The modf functions break the argument value into integral and fractional parts, ...
C11 §7.12.6.12 2

#include <math.h>

double value = 1.234;
double ipart;
double frac = modf(value, &ipart);

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A better approach for OP's need may be to first round a scaled value and then back into whole and fractional parts.

double value = 254.73;
value = round(value*100.0);
 
double frac = fmod(value, 100);  // fmod computes the floating-point remainder of x/y.
double ipart = (value - frac)/100.0;

printf("%f %f\n", ipart, frac);
254.000000 73.000000

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Ref detail: When OP uses 254.73, this is converted to the nearest float value which may be 254.729995727539....

float f = 254.73;
printf("%.30f\n", f);
// 254.729995727539062500000000000000