如何从浮点数中得到精确的小数部分作为整数？

Question

There is a lot of questions and answer about decimal & integer extraction from floating point numbers and taking output for some specific decimal points. But none could solve my problem. Please if anyone can help me to solve my problem-

I was actually 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 ie 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);

---

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

---

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

Answer 2

You can use sprintf and sscanf to print the value to a string and then extract the fraction. The %*d scans and discards the first integer of the formatted string. A dot is scanned and then the fraction.

#include <stdio.h>        

int main( void)           
{                         
    char fp[30];          
    int fraction;         
    float f = 254.73f;    

    sprintf ( fp, "%.2f", f);
    sscanf ( fp, "%*d.%d", &fraction);
    printf ( "%d\n", fraction);

    return 0;                  
}

Answer 3

The easiest way is to use standard library function ceil from <math.h> .
The float number 254.73 may be converted to 254.7299957275390625000000 .
f-integer will give 0.7299957275390625000000 .
Now multiply it by 100 and use ceil function to get the smallest integer value not less than 72.99957275390625000000 .

int fractional_part_in_integer = ((int)ceil(fractional*100)) % 100;

---

UPDATE: As pointed in a comment by @Sneftel , the above suggested method in this answer will not work consistently.

A simple hack is to use round function from math.h to round the f and then extract the fractional part

float f=254.73;

int int_part = (int)f;
float fractional = round(f*100)/100 - int_part;
int fractional_part_in_integer = (int)(fractional*100);

printf("%d, %d\n ", int_part, fractional_part_in_integer);

Output:

254, 73