Thread: 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:

 // 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

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:

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

I used this little piece of code:

Code:

#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;
}

and I got these results:

Code:

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

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.