C Examples

C Program - Count digits in an Integer



In C, count of digits in a given integer can be found out by using below mentioned methods.

Method 1: Using iteration

The method involves the following steps:

Input: MyNum
Step 1: Initialize the count = 0
Step 2: Iterate over MyNum while it is not zero.
   Step 2a: Update MyNum by MuNum / 10
   Step 2b: Increase count by 1
Step 3: Return count

Example:

Input: 564
count: 0

Iteration 1: 
    MyNum: 564 / 10 = 56
    count: 0 + 1 = 1

Iteration 2: 
    MyNum: 56 / 10 = 5
    count: 1 + 1 = 2

Iteration 3: 
    MyNum: 5 / 10 = 0
    count: 2 + 1 = 3

return count = 3

The below block of code shows the implementation of above concept:

#include <stdio.h>

//function to count number of digits
static int countDigits(long long MyNum){
  int count = 0;

  while(MyNum != 0){
    MyNum = MyNum / 10;
    count++;
  }
  return count;
}

int main() {
  long long x = 123;
  long long y = 459709;
  printf("%lli contains: %u digits \n", x, countDigits(x));
  printf("%lli contains: %u digits \n", y, countDigits(y));
  return 0;
}

The above code will give the following output:

123 contains: 3 digits
459709 contains: 6 digits

Method 2: Using Recursion

The above result can also be achieved using recursive function. Consider the example below:

#include <stdio.h>

//function to count number of digits
static int countDigits(long long MyNum) {
  if(MyNum != 0)
    return 1 + countDigits(MyNum/10);
  else
    return 0;
}

int main() {
  long long x = 564;
  long long y = 980620;
  printf("%lli contains: %u digits \n", x, countDigits(x));
  printf("%lli contains: %u digits \n", y, countDigits(y));
  return 0;
}

The above code will give the following output:

564 contains: 3 digits
980620 contains: 6 digits

Method 3: Using log function

The log (base-10) of absolute value of a given number can be used to find out the number of digits in a given number. Consider the example below:

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

//function to count number of digits
static int countDigits(long long MyNum) {
  return log10(fabs(MyNum)) + 1;
}

int main() {
  long long x = 564;
  long long y = -12345;
  long long z = 980620;
  printf("%lli contains: %u digits \n", x, countDigits(x));
  printf("%lli contains: %u digits \n", y, countDigits(y));
  printf("%lli contains: %u digits \n", z, countDigits(z));
  return 0;
}

The above code will give the following output:

564 contains: 3 digits
-12345 contains: 5 digits
980620 contains: 6 digits

Method 4: Using length of string

This can also be achieved by converting the absolute value of number into a string and then finding out the length of the string to get the number of digits in the original number. Consider the example below:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//function to count number of digits
static int countDigits(int MyNum) {
  char MyStr[50];

  //convert MyNum to string 
  sprintf(MyStr, "%i", abs(MyNum));

  return strlen(MyStr);
}

int main() {
  int x = 564;
  int y = -12345;
  int z = 980620;
  printf("%i contains: %u digits \n", x, countDigits(x));
  printf("%i contains: %u digits \n", y, countDigits(y));
  printf("%i contains: %u digits \n", z, countDigits(z));
  return 0;
}

The above code will give the following output:

564 contains: 3 digits
-12345 contains: 5 digits
980620 contains: 6 digits




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