Getting the fractional part of a double value in integer without losing precision

问题

i want to convert the fractional part of a double value with precision upto 4 digits into integer. but when i do it, i lose precision. Is there any way so that i can get the precise value?

#include<stdio.h>
int main()
{
    double number;
    double fractional_part;
    int output;
    number = 1.1234;
    fractional_part = number-(int)number;
    fractional_part = fractional_part*10000.0;
    printf("%lf\n",fractional_part);
    output = (int)fractional_part;
    printf("%d\n",output);
    return 0;
}

i am expecting output to be 1234 but it gives 1233. please suggest a way so that i can get desired output. i want the solution in C language.

回答1:

Assuming you want to get back a positive fraction even for negative values, I'd go with

(int)round(fabs(value - trunc(value)) * 1e4)

which should give you the expected result 1234.

If you do not round and just truncate the value

(int)(fabs(value - trunc(value)) * 1e4)

(which is essentially the same as your original code), you'll end up with the unexpected result 1233 as 1.1234 - 1.0 = 0.12339999999999995 in double precision.

Without using round(), you'll also get the expected result if you change the order of operations to

(int)(fabs(value * 1e4 - trunc(value) * 1e4))

If the integral part of value is large enough, floating-point inaccuracies will of course kick in again.

You can also use modf() instead of trunc() as David suggests, which is probably the best approach as far as floating point accuracy goes:

double dummy;
(int)round(fabs(modf(value, &dummy)) * 1e4)

回答2:

number= 1.1234, whole=1, fraction=1234

int main()
{
 double number;
 int whole, fraction;
 number = 1.1234;
 whole= (int)number;
 fraction =(int)(number*10000);
 fraction = fraction-(whole *10000);
 printf("%d\n",fraction);
 printf("%d\n",whole);
 return 0;
}

回答3:

A solution for any number could be:

#include <cmath>

using namespace std;

int _tmain(int argc, _TCHAR* argv[])

{ 
float number = 123.46244;
float number_final;
float temp = number;  // keep the number in a temporary variable
int temp2 = 1;        // keep the length of the fractional part

while (fmod(temp, 10) !=0)   // find the length of the fractional part
{
    temp = temp*10;
    temp2 *= 10;
}

temp /= 10;       // in tins step our number is lile this xxxx0
temp2 /= 10;
number_final = fmod(temp, temp2);

cout<<number_final;

getch();
return 0;
}

回答4:

Use modf and ceil

#include <stdio.h>
#include <math.h>

int main(void)
{
    double param, fractpart, intpart;
    int output;

    param = 1.1234;
    fractpart = modf(param , &intpart);
    output = (int)(ceil(fractpart * 10000));
    printf("%d\n", output);

    return 0;
}

来源：https://stackoverflow.com/questions/17259315/getting-the-fractional-part-of-a-double-value-in-integer-without-losing-precisio

标签

double

type-conversion

precision

floating