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public int count(int num) {
		int count = 0;

		while ((num / 10) > 0) {
			num = num / 10;
			count++;
		}
		return count + 1;

}

- Avinash December 04, 2011 | Flag Reply

Won't be the fastest variant, but the easiest for me :-)

int count(int num){
char tmp[20];
snprintf(tmp, sizeof(tmp), "%d", num);
return strlen(tmp);
}

- JeffJ January 13, 2012 | Flag Reply

<pre lang="" line="1" title="CodeMonkey34461" class="run-this">int count(int N)
{
int lenOfN = 0;
while(N > 0)
{
lenOfN++;
N = N / 10;
}
return lenOfN;
}
</pre><pre title="CodeMonkey34461" input="yes">
</pre>

- Anonymous December 03, 2011 | Flag Reply

You can use Log base 10 then add 1
Examples:
log 11 = 1.04139269 + 1 = 2 digits
log 123 = 2.08990511 + 1= 3 digits
log 1000 = 3 + 1 = 4 digits

- HopeIFindAJob December 08, 2011 | Flag Reply

try this {{{ int count (int num) { int base = 100000; int count = 6 do{ temp = num/base; if (temp==0) { base/=10; count--; continue; } if (temp !=0) { if (temp/10) { base*=10; count++;continue;} else { return count;} } } - Varun January 11, 2012 | Flag Reply

int count(int num)
{
static int i=1;
while(num)
{
num=num/10;
if(num)
i++;
}
return i;
}

- Anonymous January 19, 2012 | Flag Reply

int count(int num)
{
static int i=1;
while(num)
{
num=num/10;
if(num)
i++;
}
return i;
}

- Anonymous January 19, 2012 | Flag Reply

int count(int num)
{
static int i=1;
while(num)
{
num=num/10;
if(num)
i++;
}
return i;
}

- Rahul Mahindru January 19, 2012 | Flag Reply

int count(int num)
{
    int c = 1; 
    while((num/=10) || (num%10)) c++;  
    return c;
}

- Arijit January 31, 2012 | Flag Reply

we can use string function, but it works only on integer values and positive values.. if u want to work on negative, ignore - sign before assigning to string, or float/double values ignore value after dot.

public static void main(String[] args) {
		String number="200";
		System.out.println("number of characters in the number is: "+number.length());

}

- Dhanalakshmi November 27, 2012 | Flag Reply

public class JavaApplication1 {

    
    public static void main(String[] args) {
      int number=1233, count=1;
      while(number/10!=0){
          number=number/10;
          count=count+1;
      }
      
      System.out.println(count);
    }
}

- sonali gupta August 08, 2015 | Flag Reply

public class JavaApplication1 {

    
    public static void main(String[] args) {
      int number=1233, count=1;
      while(number/10!=0){
          number=number/10;
          count=count+1;
      }
      
      System.out.println(count);
    }
}

- sonaligupta847 August 08, 2015 | Flag Reply

Int I = 10;

For (I = 6; I >= 3; I—)
{
Printf(“%d”, I);
}

- Anonymous November 02, 2021 | Flag Reply

for(let i=0;i<100;)console.log((++i%3?'':'fizz')+(i%5?'':'buzz')||i)

- Anonymous March 22, 2022 | Flag Reply

I will use the modulo and divide operands to store the remainder and the quotient and count the characters in a number. I am assuming the question is to find no.of characters in a number like 4328.

int count ( int num)
{
   int n = 0;
   int rem = num % 10;

   while (rem != 0)
   {
       n = n + 1;
       num = num / 10;
       rem = num % 10;
   }
   return n;
}

- Anonymous December 03, 2011 | Flag Reply