Thread: is there a function that counts the digits of an int

Code:

#include <math.h>
...
(int)log10( n ) + 1;

The obvious way is to sprintf() it to a temp buffer and then use strlen

>Just curious because it would make printing out an matrix much easier.

There is also this macro I found on another list.

Code:

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

#define DECIMALS(itype) \
(sizeof(itype) * (CHAR_BIT * 12655UL) / 42039 \
+ ((itype)-1 < 0) + 1)

int main(void)
{
   int matrix[][5] = 
   {
      { INT_MIN, INT_MIN + 1, INT_MIN + 2, -1, 0 },
      { 0, 1, INT_MAX - 2, INT_MAX - 1, INT_MAX },
   };
   size_t i,j;
   for(i = 0; i < sizeof matrix / sizeof *matrix; ++i)
   {
      for(j = 0; j < sizeof *matrix / sizeof **matrix; ++j)
      {
         printf("%*d ", DECIMALS(int), matrix[i][j]);
      }
      putchar('\n');
   }
   return 0;
}

/* my output
-2147483648 -2147483647 -2147483646          -1           0 
          0           1  2147483645  2147483646  2147483647 
*/

This performs (an approximation of) the logarithmic calculation at compile time.

Speaking of macros...

Code:

#define ULEN32(x) ( \
    (x)<10?1: \
    (x)<100?2: \
    (x)<1000?3: \
    (x)<10000?4: \
    (x)<100000?5: \
    (x)<1000000?6: \
    (x)<10000000?7: \
    (x)<100000000?8: \
    (x)<1000000000?9: 10 )

Quzah.

Many suggestions so far. Oh well, here's another:

Code:

int GetNrOfDigits(int Number)
{
   int Digits = 0;
   while(Number > 0)
   {
      Digits++;
      Number /= 10;
   }
   return Digits;
}

(Only works for positive integers, and 0 comes out as 0 digits though, you could use a special case if this is unwanted)