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int any_integer = 54321
// Add one to base 10 of the number (i.e. log10(1234) + 1 = 3. + 1 = 4 digits
int numberOfDigits = log10(any_integer) + 1
// Alternate
int numberOfDigits = ceil(log10(any_integer))
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digits = (number == 0) ? 1 : (log10(number) + 1);
//or
while (number > 0)
{
number /= 10;
digits++;
}
//see: https://ideone.com/P1h8Ne
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/* C Program to Count Number of Digits in a Number using While Loop */
#include <stdio.h>
int main()
{
int Number, Reminder, Count=0;
printf("\n Please Enter any number\n");
scanf("%d", &Number);
while(Number > 0)
{
Number = Number / 10;
Count = Count + 1;
}
printf("\n Number of Digits in a Given Number = %d", Count);
return 0;
}
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#include <stdio.h>
int main(){
int num = 1024;
while(num != 0){
int digit = num % 10;
num = num / 10;
printf("%d\n", digit);
}
return 0;
}
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#include<stdio.h>
int main()
{
int n,d;
printf("enter no.:");
scanf("%d",&n);
while(n != 0)
{
d=n%10;
n=n/10;
printf("%d\n",d);
}
return 0;
}