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).

1. After first iteration, the value of n will be 345 and the count is incremented to 1.
2. After second iteration, the value of n will be 34 and the count is incremented to 2.
3. After third iteration, the value of n will be 3 and the count is incremented to 3.
4. 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 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    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



Output :

Number of digits : 9

Recursive Solution:

C++

 // Recursive C++ program to count number of // digits in a number #include 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    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



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 #include    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



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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