How do I determine the number of decimal digits of an integer in C?

Question

for instance,

n = 3432, result 4

n = 45, result 2

n = 33215, result 5

n = -357, result 3

I guess I could just turn it into a string then get the length of the string but that seems convoluted and hack-y.

Answer 1

Divide by 10 in a loop until the result reaches zero. The number of iterations will correspond to the number of decimal digits.

Assuming that you expect to get 0 digits in a zero value:

int countDigits( int value )
{
    int result = 0;
    while( value != 0 ) {
       value /= 10;
       result++;
    }
    return result;
}

Answer 2

floor (log10 (abs (x))) + 1

http://en.wikipedia.org/wiki/Logarithm

Answer 3

The recursive approach :-)

int numPlaces (int n) {
    if (n < 0) return numPlaces ((n == MININT) ? MAXINT : -n);
    if (n < 10) return 1;
    return 1 + numPlaces (n / 10);
}

Or iterative:

int numPlaces (int n) {
    int r = 1;
    if (n < 0) n = (n == MININT) ? MAXINT : -n;
    while (n > 9) {
        n /= 10;
        r++;
    }
    return r;
}

Or raw speed:

int numPlaces (int n) {
    if (n < 0) n = (n == MININT) ? MAXINT : -n;
    if (n < 10) return 1;
    if (n < 100) return 2;
    if (n < 1000) return 3;
    if (n < 10000) return 4;
    if (n < 100000) return 5;
    if (n < 1000000) return 6;
    if (n < 10000000) return 7;
    if (n < 100000000) return 8;
    if (n < 1000000000) return 9;
    /*      2147483647 is 2^31-1 - add more ifs as needed
       and adjust this final return as well. */
    return 10;
}

Those above have been modified to better process MININT. On any weird systems that don't follow sensible 2ⁿ two's complement rules for integers, they may need further adjustment.

The raw speed version actually outperforms the floating point version, modified below:

int numPlaces (int n) {
    if (n == 0) return 1;
    return floor (log10 (abs (n))) + 1;
}

With a hundred million iterations, I get the following results:

Raw speed with 0:            0 seconds
Raw speed with 2^31-1:       1 second
Iterative with 2^31-1:       5 seconds
Recursive with 2^31-1:       6 seconds
Floating point with 1:       6 seconds
Floating point with 2^31-1:  7 seconds

That actually surprised me a little - I thought the Intel chips had a decent FPU but I guess general FP operations still can't compete with hand-optimized integer code.

Update following stormsoul's suggestions:

Testing the multiply-iterative solution by stormsoul gives a result of 4 seconds so, while it's much faster than the divide-iterative solution, it still doesn't match the optimized if-statement solution.

Choosing the arguments from a pool of 1000 randomly generated numbers pushed the raw speed time out to 2 seconds so, while it appears there may have been some advantage to having the same argument each time, it's still the fastest approach listed.

Compiling with -O2 improved the speeds but not the relative positions (I increased the iteration count by a factor of ten to check this).

Any further analysis is going to have to get seriously into the inner workings of CPU efficiency (different types of optimization, use of caches, branch prediction, which CPU you actually have, the ambient temperature in the room and so on) which is going to get in the way of my paid work :-). It's been an interesting diversion but, at some point, the return on investment for optimization becomes too small to matter. I think we've got enough solutions to have answered the question (which was, after all, not about speed).

Further update:

This will be my final update to this answer barring glaring errors that aren't dependent on architecture. Inspired by stormsoul's valiant efforts to measure, I'm posting my test program (modified as per stormsoul's own test program) along with some sample figures for all methods shown in the answers here. Keep in mind this is on a particular machine, your mileage may vary depending on where you run it (which is why I'm posting the test code).

Do with it as you wish:

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

#define numof(a) (sizeof(a) / sizeof(a[0]))

/* Random numbers and accuracy checks. */

static int rndnum[10000];
static int rt[numof(rndnum)];

/* All digit counting functions here. */

static int count_recur (int n) {
    if (n < 0) return count_recur ((n == INT_MIN) ? INT_MAX : -n);
    if (n < 10) return 1;
    return 1 + count_recur (n / 10);
}

static int count_diviter (int n) {
    int r = 1;
    if (n < 0) n = (n == INT_MIN) ? INT_MAX : -n;
    while (n > 9) {
        n /= 10;
        r++;
    }
    return r;
}

static int count_multiter (int n) {
    unsigned int num = abs(n);
    unsigned int x, i;
    for (x=10, i=1; ; x*=10, i++) {
        if (num < x)
            return i;
        if (x > INT_MAX/10)
            return i+1;
    }
}

static int count_ifs (int n) {
    if (n < 0) n = (n == INT_MIN) ? INT_MAX : -n;
    if (n < 10) return 1;
    if (n < 100) return 2;
    if (n < 1000) return 3;
    if (n < 10000) return 4;
    if (n < 100000) return 5;
    if (n < 1000000) return 6;
    if (n < 10000000) return 7;
    if (n < 100000000) return 8;
    if (n < 1000000000) return 9;
    /*      2147483647 is 2^31-1 - add more ifs as needed
    and adjust this final return as well. */
    return 10;
}

static int count_revifs (int n) {
    if (n < 0) n = (n == INT_MIN) ? INT_MAX : -n;
    if (n > 999999999) return 10;
    if (n > 99999999) return 9;
    if (n > 9999999) return 8;
    if (n > 999999) return 7;
    if (n > 99999) return 6;
    if (n > 9999) return 5;
    if (n > 999) return 4;
    if (n > 99) return 3;
    if (n > 9) return 2;
    return 1;
}

static int count_log10 (int n) {
    if (n < 0) n = (n == INT_MIN) ? INT_MAX : -n;
    if (n == 0) return 1;
    return floor (log10 (n)) + 1;
}

static int count_bchop (int n) {
    int r = 1;
    if (n < 0) n = (n == INT_MIN) ? INT_MAX : -n;
    if (n >= 100000000) {
        r += 8;
        n /= 100000000;
    }
    if (n >= 10000) {
        r += 4;
        n /= 10000;
    }
    if (n >= 100) {
        r += 2;
        n /= 100;
    }
    if (n >= 10)
        r++;

    return r;
}

/* Structure to control calling of functions. */

typedef struct {
    int (*fnptr)(int);
    char *desc;
} tFn;

static tFn fn[] = {
    NULL,                              NULL,
    count_recur,    "            recursive",
    count_diviter,  "     divide-iterative",
    count_multiter, "   multiply-iterative",
    count_ifs,      "        if-statements",
    count_revifs,   "reverse-if-statements",
    count_log10,    "               log-10",
    count_bchop,    "          binary chop",
};
static clock_t clk[numof (fn)];

int main (int c, char *v[]) {
    int i, j, k, r;
    int s = 1;

    /* Test code:
        printf ("%11d %d\n", INT_MIN, count_recur(INT_MIN));
        for (i = -1000000000; i != 0; i /= 10)
            printf ("%11d %d\n", i, count_recur(i));
        printf ("%11d %d\n", 0, count_recur(0));
        for (i = 1; i != 1000000000; i *= 10)
            printf ("%11d %d\n", i, count_recur(i));
        printf ("%11d %d\n", 1000000000, count_recur(1000000000));
        printf ("%11d %d\n", INT_MAX, count_recur(INT_MAX));
    /* */

    /* Randomize and create random pool of numbers. */

    srand (time (NULL));
    for (j = 0; j < numof (rndnum); j++) {
        rndnum[j] = s * rand();
        s = -s;
    }
    rndnum[0] = INT_MAX;
    rndnum[1] = INT_MIN;

    /* For testing. */
    for (k = 0; k < numof (rndnum); k++) {
        rt[k] = (fn[1].fnptr)(rndnum[k]);
    }

    /* Test each of the functions in turn. */

    clk[0] = clock();
    for (i = 1; i < numof (fn); i++) {
        for (j = 0; j < 10000; j++) {
            for (k = 0; k < numof (rndnum); k++) {
                r = (fn[i].fnptr)(rndnum[k]);
                /* Test code:
                    if (r != rt[k]) {
                        printf ("Mismatch error [%s] %d %d %d %d\n",
                            fn[i].desc, k, rndnum[k], rt[k], r);
                        return 1;
                    }
                /* */
            }
        }
        clk[i] = clock();
    }

    /* Print out results. */

    for (i = 1; i < numof (fn); i++) {
        printf ("Time for %s: %10d\n", fn[i].desc, (int)(clk[i] - clk[i-1]));
    }

    return 0;
}

And here's the leaderboard for my environment:

Optimization level 0:

Time for reverse-if-statements:       1704
Time for         if-statements:       2296
Time for           binary chop:       2515
Time for    multiply-iterative:       5141
Time for      divide-iterative:       7375
Time for             recursive:      10469
Time for                log-10:      26953

Optimization level 3:

Time for         if-statements:       1047
Time for           binary chop:       1156
Time for reverse-if-statements:       1500
Time for    multiply-iterative:       2937
Time for      divide-iterative:       5391
Time for             recursive:       8875
Time for                log-10:      25438

Answer 4

Binary search pseudo algorithm to get no of digits of r in v..

if (v < 0 ) v=-v;

r=1;

if (v >= 100000000)
{
  r+=8;
  v/=100000000;
}

if (v >= 10000) {
    r+=4;
    v/=10000;
}

if (v >= 100) {
    r+=2;
    v/=100;
}

if( v>=10)
{
    r+=1;
}

return r;

Answer 5

You can do: floor (log10 (abs (x))) + 1 Or if you want to save on cycles you could just do comparisons

if(x<10)
  return 1;
if(x<100)
  return 2;
if(x<1000)
  return 3;
etc etc

This avoids any computationally expensive functions such as log or even multiplication or division. While it is inelegant this can be hidden by encapsulating it into a function. It isn't complex or difficult to maintain so I would not dismiss this approach on account of poor coding practice; I feel to do so would be throwing the baby out with the bath water.

Answer 6

int n = 437788; int N = 1; while (n /= 10) N++;

Answer 7

if (x == MININT) return 10;  //  abs(MININT) is not defined
x = abs (x);
if (x<10) return 1;
if (x<100) return 2;
if (x<1000) return 3;
if (x<10000) return 4;
if (x<100000) return 5;
if (x<1000000) return 6;
if (x<10000000) return 7;
if (x<100000000) return 8;
if (x<1000000000) return 9;
return 10; //max len for 32-bit integers

Very inelegant. But quicker than all the other solutions. Integer Division and FP logs are expensive to do. If performance isn't an issue, the log10 solution is my favorite.

Answer 8

From Bit Twiddling Hacks :

Find integer log base 10 of an integer the obvious way

Note the ordering of comparisons in it.

Answer 9

DON'T use floor(log10(...)). These are floating-point functions, and slow ones, to add. I believe the fastest way would be this function:

int ilog10(int num)
{
   unsigned int num = abs(num);
   unsigned int x, i;
   for(x=10, i=1; ; x*=10, i++)
   {
      if(num < x)
         return i;
      if(x > INT_MAX/10)
         return i+1;
   }
}

Note that the binary search version some people suggested could be slower due to branch mispredictions.

EDIT:

I did some testing, and got some really interesting results. I timed my function together with all the functions tested by Pax, AND the binary search function given by lakshmanaraj. The testing is done by the following code snippet:

start = clock();
for(int i=0; i<10000; i++)
   for(int j=0; j<10000; j++)
      tested_func(numbers[j]);
end = clock();
tested_func_times[pass] = end-start;

Where the numbers[] array contains randomly generated numbers over the entire range of the int type (barring MIN_INT). The testing was repeated for each tested function on THE SAME numbers[] array. The entire test was made 10 times, with results averaged over all passes. The code was compiled with GCC 4.3.2 with -O3 optimization level.

Here are the results:

floating-point log10:     10340ms
recursive divide:         3391ms
iterative divide:         2289ms
iterative multiplication: 1071ms
unrolled tests:           859ms
binary search:            539ms

I must say I got really astonished. The binary search performed far better than I thought it would. I checked out how GCC compiled this code to asm. O_O. Now THIS is impressive. It got optimized much better than I thought possible, avoiding most branches in really clever ways. No wonder it is FAST.