Changing float numbers to int numbers in c

Question

In some part of my program, I have an array which contains float numbers (for example 9.8, 5.0, 4.45). What I have to do is take those numbers to another array and if they are float (I mean if they have digits after the decimal point that are not zero), then multiply them by 10 enough times to make it an int value (so from the first example, after that I should have 98, 5, 445). I know it doesn't sound clear but it's hard for me to describe my problem better.

My attempt was to do this but I have errors.

for(i=0;i<size;i++)
    {
       if(teb[i]%10!=0)
       {
           while(teb[i]%1!=0)
           {
               teb[i]=teb[i]*10;
           }
       }
    }

I have error: invalid operator to binary % (have float and int) and I am not sure if I can use char or similar variables.

Answer 1

Normally I don't provide answers to homework problems, but since this is such a bad problem, I'm making an exception.

Before I proceed, let me say that the original problem is meaningless and impossible. Since it's impossible, there's no good way to solve it. So the program I'm going to present has a number of problems, some quite serious.

The core of the program does what the problem asks: As long as the fractional part of one of the numbers is not 0, it multiplies the number by 10. Computing "the fractional part of the number" is performed by an auxiliary function I've written, fractionalpart(). First we'll look at the program, then I'll explain its remaining problems.

#include <stdio.h>

float in[] = {9.8, 5.0, 4.45};
int out[3];

float fractionalpart(float f)
{
    return f - (int)f;
}

int main()
{
    int i;
    for(i = 0; i < 3; i++) {
        float tmp = in[i];
        while(fractionalpart(tmp) > 0.001) {
            tmp = tmp * 10;
        }

        out[i] = tmp;
    }

    printf("out:\n");
    for(i = 0; i < 3; i++) printf("%d\n", out[i]);
}

The problems this has relate to the fact that, on an ordinary computer using ordinary floating-point representations, there is no such number as 9.8. There's no such number as 4.45, either. What you thought was the float number 4.45, for example, is represented internally as something like 4.44999980926513671875. So a "proper" version of this program would multiply it by 10 a total of twenty times, converting it to 444999980926513671875. But that number won't even fit in a 64-bit integer!

So this program cheats: it doesn't loop until the fractional part is exactly 0; it instead loops until the fractional part is less than 0.001. Where did that number come from? The answer is: nowhere, I pulled it out of the air, it seemed like a good guess. You can experiment with bigger or smaller fudge factors if you like.

The bigger problem -- and you've probably thought of this already -- is that the program assumes that the "fractional part" actually does go down to something like 0.001 or 0.0001. But that's not necessarily true! As we just saw, the number 4.45 is "really" 4.44999980926513671875 inside, so after multiplying by 10 twice, it looks like the program is going to have an integer value of 444 (not 445), and the fractional part is going to be 0.999980926513671875, which is not less than 0.001, so the program is going to keep going!

Actually, it doesn't keep going, it does get the "right answer", at least on my machine (and I think I know why), but there are probably plenty of numbers (roughly half of them, actually) where there will be an intermediate result more like 0.999 than 0.001, and this program will do the wrong thing.

I hate posting code with serious problems like this, it's very tempting to try to write an "improved" version, but it would be much messier, and really, the original question is such a horrible one that it just isn't worth it. (It would be instructive to figure out and explain why this program does seems to work, at least most of the time -- why it doesn't get confused by the occasional 0.999 value -- but I don't have time for that just now, either.)

Answer 2

Most decimal fractions expressed as some finite number of digits in base 10 cannot be stored exactly in the binary floating point representations typically used in most implementations.

If you know the numbers are only supposed to be accurate to a fixed number of decimal places n, you can first multiply by 10ⁿ and round to the nearest integer. Then strip off up to n trailing zero digits.

For example, for up to 3 decimal places:

for(i=0;i<size;i++)
{
   long x = round(teb[i] * 1000.0);
   for (int j = 0; j < 3; j++)
   {
       if (x % 10 == 0)
       {
           x /= 10;
       }
       else
       {
           break;
       }
   }
   teb[i] = x;
}

Answer 3

In my opinion, you'll not be able to manage this sanely without using string formatting and then converting from string back to floating-point. I use double rather than float. This is the code I came up with, including your three test values as the first three.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

static inline int all_same_as(char *str, char c)
{
    while (*str != '\0')
    {
        if (*str++ != c)
            return 0;
    }
    return 1;
}

static inline void zap_trailing(char *str, char c)
{
    char *end = str + strlen(str);
    while (end > str)
    {
        if (*(end - 1) != c)
            return;
        *--end = '\0';
    }
}

static inline void zap_leading(char *str, char c)
{
    char *end = str;
    while (*end == c)
        end++;
    if (end > str)
        memmove(str, end, strlen(end) + 1);  /* Not memcpy()! */
}

int main(void)
{
    double d[] =
    {
        9.8,
        5.0,
        4.45,
        1.0,
        18.0,
        9.25,
        8.719,
        2.78128,
        2721.0/1001.0,
        3.14159,
        355.0/113.0,
        0.001234,   /* Smaller */
        1.2345E+13, /* Biggest */
        1.2345E-13, /* Smallest */
    };
    enum { NUM_D = sizeof(d) / sizeof(d[0]) };

    for (int i = 0; i < NUM_D; i++)
    {
        char buffer1[32];
        snprintf(buffer1, sizeof(buffer1), "%.6f", d[i]);
        char buffer2[32];
        strcpy(buffer2, buffer1);
        zap_leading(buffer2, ' ');              /* Not needed with %.6f; was needed with %10.6f */
        char *dot = strchr(buffer2, '.');
        if (all_same_as(dot + 1, '0'))
            *dot = '\0';
        else
        {
            zap_trailing(dot + 1, '0');
            size_t len = strlen(dot + 1);
            memmove(dot, dot + 1, len + 1);     /* Not memcpy()! */
        }
        double iv = strtod(buffer2, NULL);
        printf("%8g = %8.2g = %8.3g = %10s (converted: %10s - [%s] %g)\n",
               d[i], d[i], d[i], buffer1, buffer2, buffer2, iv);
    }
    return 0;
}

Output:

     9.8 =      9.8 =      9.8 =   9.800000 (converted:         98 - [98] 98)
       5 =        5 =        5 =   5.000000 (converted:          5 - [5] 5)
    4.45 =      4.5 =     4.45 =   4.450000 (converted:        445 - [445] 445)
       1 =        1 =        1 =   1.000000 (converted:          1 - [1] 1)
      18 =       18 =       18 =  18.000000 (converted:         18 - [18] 18)
    9.25 =      9.2 =     9.25 =   9.250000 (converted:        925 - [925] 925)
   8.719 =      8.7 =     8.72 =   8.719000 (converted:       8719 - [8719] 8719)
 2.78128 =      2.8 =     2.78 =   2.781280 (converted:     278128 - [278128] 278128)
 2.71828 =      2.7 =     2.72 =   2.718282 (converted:    2718282 - [2718282] 2.71828e+06)
 3.14159 =      3.1 =     3.14 =   3.141590 (converted:     314159 - [314159] 314159)
 3.14159 =      3.1 =     3.14 =   3.141593 (converted:    3141593 - [3141593] 3.14159e+06)
0.001234 =   0.0012 =  0.00123 =   0.001234 (converted:    0001234 - [0001234] 1234)
1.2345e+13 =  1.2e+13 = 1.23e+13 = 12345000000000.000000 (converted: 12345000000000 - [12345000000000] 1.2345e+13)
1.2345e-13 =  1.2e-13 = 1.23e-13 =   0.000000 (converted:          0 - [0] 0)

You can make choices about how many decimal digits to support, etc. I chose to use up to 6 after the decimal point (format %.6f used with snprintf()). I included very small (1.23E-13) and very large (1.23E+13) values; the behaviour for even bigger or smaller values is similar.

I initially used %10.6f in the snprintf() statement. When I did that, values were printed with leading blanks. The zap_leading() function removes those. The code has since been revised to use %.6f and there are no leading blanks to zap. I left the code to zap leading characters in place; it could be removed too.