Potentially useful code

I figured out a way to calculate the max number of digits in a base 10 integer using only the number of bits in said integer, here's my code:

Code:

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// Calculate the number of base 10 digits #ifndef INT_BASE10_DIG #define INT_BASE10_DIG(BITS) (((((BITS)%3)==0)?(BITS):((BITS)+(3-((BITS)%3))))/3) #endif

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I figured it out after looking at the numbers on a PC calculator in scientific binary mode and was appending 1s slowly and saw that after the total 1s moved above a divisor of 3 the number of decimal digits would go up. I gave the thread this title in case others wanted to add their own to it and end up with being a sticky thread edit: This can go in the same #ifdef to caculate the needed integer size

Code:

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#define BASE10_INT_BIT(DIGS) ((DIGS)*3)+(CHAR_BIT-(((DIGS)*3)%CHAR_BIT))

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I used this little piece of code:

Code:

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#include <stdio.h>
#include <limits.h>

#define BASE10_INT_BIT(DIGS) ((DIGS)*3)+(CHAR_BIT-(((DIGS)*3)%CHAR_BIT))

int main(void)
{
        int i;

        printf("Binary | Decimal\n");

        for (i = 1; i <= 16; i++) {
                printf("%4d    %4d\n", i, BASE10_INT_BIT(i));
        }

        return 0;
}

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and I got these results:

Code:

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Binary | Decimal
  1        8
  2        8
  3      16
  4      16
  5      16
  6      24
  7      24
  8      32
  9      32
  10      32
  11      40
  12      40
  13      40
  14      48
  15      48
  16      56

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Those numbers don't seem right at all, unless I misunderstood what you're calculating.

EDIT: Is it backwards? I think I used it backwards. That macro seems to take the number of decimal digits and converts it into bits...

The actual number you need is given by log (10)÷log(2)
Which is approximately 3.32 binary bits for each decimal digit.

So if your approximation can manage that, you should be good :)

Don't forget the +1 in your approximation for storing a \0.