Digits digits numbers without C ++ loop

Question

I need to count the number of digits in a number.

I divide the number to 10 until I get 0. Each iteration increments the counter.

int num;
cin>>num;  
while(num > 0)  
{  
  counter++;
  num = num / 10;   
}

The problem is not using loops or recursion, but simply the expression if.

Is it possible?

Answer 1

counter = log (num) / log (10)

in C ++:

#include <cmath>
....
counter = num == 0 ? 1 : log10(std::abs(num)) + 1;

what you want is a log feature.

cplusplus - log10

cplusplus - std :: abs

Answer 2

An easy way, albeit somewhat expensive, turn your number into a string and take its size, as shown below:

#include <iostream>
#include <string>

int main() {
  int i = 1232323223;
  std::string str = std::to_string(std::abs(i));
  std::cout << "Number of Digits: " << str.size() <<std::endl;
}

LIVE DEMO

Answer 3

- sprintf, :

int digits(int n)
{
    char s[32];
    int len = sprintf(s, "%d", n);
    if (n < 0) len--; // NB: handle negative case
    return len;
}

Answer 4

, , - ( 32- ):

int numDigitsU (unsigned int 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;
    /*      4294967295 is 2^32-1 - add more ifs as needed
       and adjust this final return as well. */
    return 10;
}

, , MININT:

int numDigitsS (int n) {
    if (n == MININT) n = MAXINT;  // same number of digits, usually.
    if (n < 0) n = -n;            // reverse sign.
    return numDigitsU (n);        // call the unsigned variant.
}

unsigned int.

: , /.

Answer 5

, , int . , 10, 100, 1000 .. .

int num = abs(number);
if (num < 10000)
{
    if (num < 100)
        return num < 10 ? 1:2;
    else
        return num < 1000 ? 3:4;   
}
else
{
    ...
}