Thread: Floating Point Not Rounding

Hum, you got me there. I don't know the reason for that output. Someone else might, though. Maybe some systems round the digit 5 down instead of up.

printf follows the more "statistically correct" way of rounding 5 to the nearest even number. As a result, if you want to round a 5, it depends whether the second digit left of it is even or odd( left of the one you round to ). If it is odd, you round down, if even you round up.

printf follows the more "statistically correct" way of rounding 5 to the nearest even number. As a result, if you want to round a 5, it depends whether the second digit left of it is even or odd( left of the one you round to ). If it is odd, you round down, if even you round up.

O_o

The situation shown in the example doesn't have much to do with any tie-breaking aspects of the rounding method. The values `1.9375' and `1.890625' once rounded, with respect to the relevant `5' digit value, would show (`1.938' and `1.89062') the tie-breaking aspect of the rounding method. The values in the situation shown doesn't have an exact representation so no tie-breaking is necessary.

The problem, of sorts, is how floating-point values are stored in binary.

The value `9.995' is most likely `~9.9949998' which is just being rounded down.

Soma

I was really interested in the explanations and made some tests...

Code:

#include<stdio.h>

int main()
{
    float f;
    double d;
    scanf("%f",&f);
    d = f;
    printf("%f %e %.2f\n",f,f,f );
    printf("%f %e %.2f\n",d,d,d );
    scanf("%f",&f);
    d = f;
    printf("%f %e %.2f\n",f,f,f );
    printf("%f %e %.2f\n",d,d,d );
    return 0;
}

And got following results

9.995
9.995000 9.995000e+00 9.99
9.995000 9.995000e+00 9.99
9.985
9.985000 9.985000e+00 9.98
9.985000 9.985000e+00 9.98

look to me 5 is being rounded down no matter what...

I was really interested in the explanations and made some tests...

O_o

Your test is flawed. Your results are irrelevant.

The values `9.995' and `9.985' do not exist in IEEE754 binary, arguably the most common, representation.

Code:

#include <stdio.h>

void Dump
(
    float f
)
{
    char s[16];
    int c = 2;
    do {
        sprintf(s, "%%.%df\n", c);
        printf(s, f);
    } while(c++ < 9);
}

int main()
{
    float s;
    while(1 == scanf("%f", &s))
    {
        Dump(s);
    }
    return(0);
}

Code:

10.015
10.005
9.995
9.985
9.975

Code:

10.02
10.015
10.0150
10.01500
10.015000
10.0150003
10.01500034
10.015000343
10.01
10.005
10.0050
10.00500
10.005000
10.0050001
10.00500011
10.005000114
9.99
9.995
9.9950
9.99500
9.995000
9.9949999
9.99499989
9.994999886
9.98
9.985
9.9850
9.98500
9.985000
9.9849997
9.98499966
9.984999657
9.98
9.975
9.9750
9.97500
9.975000
9.9750004
9.97500038
9.975000381

look to me 5 is being rounded down no matter what...

o_O

You do not have a five.

You have a five and a bit, or you have a four and a lot.

If you want to see tie-breaking applied, you must use values which can be exactly represented.

Code:

#include <stdio.h>

int main()
{
    printf("%.4f\n", 1.40625f); // down
    printf("%.5f\n", 1.421875f); // up
    printf("%.5f\n", 1.453125f); // down
    printf("%.6f\n", 1.4609375f); // up
    return(0);
}

Strangely enough, the example I've provided may also fail to show tie-breaking behavior because not all compilers have enough precision to accurately encode values of the relevant precision.

Soma

O_o

I am used to C++ so wasn't thinking about the `float'->`double' passing or the type of comments.

Soma