Program to count digits in an integer (4 Different Methods)

Count the number of digits in a long integer entered by a user.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple Iterative Solution
The integer entered by the user is stored in variable n. Then the while loop is iterated until the test expression n != 0 is evaluated to 0 (false).

After first iteration, the value of n will be 345 and the count is incremented to 1.
After second iteration, the value of n will be 34 and the count is incremented to 2.
After third iteration, the value of n will be 3 and the count is incremented to 3.
At the start of fourth iteration, the value of n will be 0 and the loop is terminated.

Then the test expression is evaluated to false and the loop terminates.

C++

// Iterative C++ program to count 
// number of digits in a number 
#include <bits/stdc++.h> 
using namespace std; 
  
int countDigit(long long n) 
{ 
    int count = 0; 
    while (n != 0) { 
        n = n / 10; 
        ++count; 
    } 
    return count; 
} 
  
// Driver code 
int main(void) 
{ 
    long long n = 345289467; 
    cout << "Number of digits : "
         << countDigit(n); 
    return 0; 
} 
  
// This code is contributed 
// by Akanksha Rai 

C

// Iterative C program to count number of 
// digits in a number 
#include <stdio.h> 
  
int countDigit(long long n) 
{ 
    int count = 0; 
    while (n != 0) { 
        n = n / 10; 
        ++count; 
    } 
    return count; 
} 
  
// Driver code 
int main(void) 
{ 
    long long n = 345289467; 
    printf("Number of digits : %d", 
           countDigit(n)); 
    return 0; 
} 

Java

// JAVA Code to count number of 
// digits in an integer 
class GFG { 
  
    static int countDigit(long n) 
    { 
        int count = 0; 
        while (n != 0) { 
            n = n / 10; 
            ++count; 
        } 
        return count; 
    } 
  
    /* Driver program to test above function */
    public static void main(String[] args) 
    { 
        long n = 345289467; 
        System.out.print("Number of digits : " + countDigit(n)); 
    } 
} 
// This code is contributed by Arnav Kr. Mandal. 

Python3

# Iterative Python program to count 
# number of digits in a number 
  
def countDigit(n): 
    count = 0
    while n != 0: 
        n //= 10
        count+= 1
    return count 
  
# Driver Code     
n = 345289467
print ("Number of digits : % d"%(countDigit(n))) 
  
# This code is contributed by Shreyanshi Arun 

C#

// C# Code to count number of 
// digits in an integer 
using System; 
  
class GFG { 
  
    static int countDigit(long n) 
    { 
        int count = 0; 
        while (n != 0) { 
            n = n / 10; 
            ++count; 
        } 
        return count; 
    } 
  
    /* Driver program to test 
    above function */
    public static void Main() 
    { 
        long n = 345289467; 
        Console.WriteLine("Number of"
                          + " digits : " + countDigit(n)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// Iterative PHP program to count  
// number of digits in a number 
  
function countDigit($n) 
{ 
    $count = 0; 
    while ($n != 0)  
    { 
        $n = round($n / 10); 
        ++$count; 
    } 
    return $count; 
} 
  
// Driver code 
$n = 345289467; 
echo "Number of digits : "
        . countDigit($n); 
  
//This code is contributed by mits 
?> 

Output :

Number of digits : 9

Recursive Solution:

C++

// Recursive C++ program to count number of 
// digits in a number 
#include <bits/stdc++.h> 
using namespace std; 
  
int countDigit(long long n) 
{ 
    if (n == 0) 
        return 0; 
    return 1 + countDigit(n / 10); 
} 
  
// Driver code 
int main(void) 
{ 
    long long n = 345289467; 
    cout << "Number of digits :" << countDigit(n); 
    return 0; 
} 
// This code is contributed by Mukul Singh. 

C

// Recursive C program to count number of 
// digits in a number 
#include <stdio.h> 
  
int countDigit(long long n) 
{ 
    if (n == 0) 
        return 0; 
    return 1 + countDigit(n / 10); 
} 
  
// Driver code 
int main(void) 
{ 
    long long n = 345289467; 
    printf("Number of digits : %d", 
           countDigit(n)); 
    return 0; 
} 

Java

// JAVA Code to count number of 
// digits in an integer 
import java.util.*; 
  
class GFG { 
  
    static int countDigit(long n) 
    { 
        if (n == 0) 
            return 0; 
        return 1 + countDigit(n / 10); 
    } 
  
    /* Driver program to test above function */
    public static void main(String[] args) 
    { 
        long n = 345289467; 
        System.out.print("Number of digits : " + countDigit(n)); 
    } 
} 
  
// This code is contributed by Arnav Kr. Mandal. 

Python3

# Recursive Python program to count 
# number of digits in a number 
  
def countDigit(n): 
    if n == 0: 
        return 0
    return 1 + countDigit(n // 10) 
  
# Driver Code     
n = 345289467
print ("Number of digits : % d"%(countDigit(n))) 
  
# This code is contributed by Shreyanshi Arun 

C#

// C# Code to count number of 
// digits in an integer 
using System; 
  
class GFG { 
  
    static int countDigit(long n) 
    { 
        if (n == 0) 
            return 0; 
        return 1 + countDigit(n / 10); 
    } 
  
    /* Driver program to test above 
    function */
    public static void Main() 
    { 
        long n = 345289467; 
        Console.WriteLine("Number of "
                          + "digits : " + countDigit(n)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// Recursive PHP program to count 
// number of digits in a number 
  
function countDigit($n) 
{ 
    if ($n == 0) 
        return 0; 
    return 1 + countDigit((int)($n / 10)); 
} 
  
// Driver Code  
$n = 345289467; 
print ("Number of digits : " .  
            (countDigit($n))); 
  
// This code is contributed by mits 
?> 

Output :

Number of digits : 9

Log based Solution:
We can use log10(logarithm of base 10) to count the number of digits of positive numbers (logarithm is not defined for negative numbers).

Digit count of N = upper bound of log10(N).

C

// Log based C program to count number of 
// digits in a number 
#include <math.h> 
#include <stdio.h> 
  
int countDigit(long long n) 
{ 
    return floor(log10(n) + 1); 
} 
  
// Driver code 
int main(void) 
{ 
    long long n = 345289467; 
    printf("Number of digits : %d", 
           countDigit(n)); 
    return 0; 
} 

Java

// JAVA Code to count number of 
// digits in an integer 
import java.util.*; 
  
class GFG { 
  
    static int countDigit(long n) 
    { 
        return (int)Math.floor(Math.log10(n) + 1); 
    } 
  
    /* Driver program to test above function */
    public static void main(String[] args) 
    { 
        long n = 345289467; 
        System.out.print("Number of digits : " + countDigit(n)); 
    } 
} 
// This code is contributed by Arnav Kr. Mandal. 

Python3

# Log based Python program to count number of 
# digits in a number 
  
# function to import ceil and log  
import math 
  
def countDigit(n): 
    return math.floor(math.log(n, 10)+1) 
  
# Driver Code 
n = 345289467
print ("Number of digits : % d"%(countDigit(n))) 
  
# This code is contributed by Shreyanshi Arun 

C#

// C# Code to count number of 
// digits in an integer 
using System; 
  
class GFG { 
  
    static int countDigit(long n) 
    { 
        return (int)Math.Floor(Math.Log10(n) + 1); 
    } 
  
    /* Driver program to test above function */
    public static void Main() 
    { 
        long n = 345289467; 
        Console.WriteLine("Number of digits : "
                          + countDigit(n)); 
    } 
} 
  
// This code is contributed by anuj_67.. 

PHP

<?php 
// Log based php program to  
// count number of digits 
// in a number 
  
function countDigit($n) 
{ 
    return floor(log10($n)+1); 
} 
  
// Driver code 
$n = 345289467; 
echo "Number of digits : ", 
    countDigit($n); 
  
// This code is contributed by ajit 
?> 

Output :

Number of digits : 9

Method 4:
We can convert the number into string and then find the length of the string to get the number of digits in the original number.

# Python3 implementation of the approach 
def count_digits(n): 
    n = str(n) 
    return len(n) 
  
# Driver code 
n = 456533457776
print(count_digits(n)) 

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

C++

C

Java

Python3

C#

PHP

C

Java

Python3

C#

PHP

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