Count the number of digits in a long integer entered by a user.
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++
#include <bits/stdc++.h>
using namespace std;
int countDigit(long long n)
{
int count = 0;
while (n != 0)
{
n = n / 10;
++count;
}
return count;
}
int main(void)
{
long long n = 345289467;
cout << "Number of digits : " << countDigit(n);
return 0;
}
|
C
#include <stdio.h>
int countDigit(long long n)
{
int count = 0;
while (n != 0)
{
n = n / 10;
++count;
}
return count;
}
int main(void)
{
long long n = 345289467;
printf("Number of digits : %d", countDigit(n));
return 0;
}
|
Java
class GFG {
static int countDigit(long n)
{
int count = 0;
while (n != 0) {
n = n / 10;
++count;
}
return count;
}
public static void main(String[] args)
{
long n = 345289467;
System.out.print("Number of digits : "
+ countDigit(n));
}
}
|
Python3
def countDigit(n):
count = 0
while n != 0:
n //= 10
count += 1
return count
n = 345289467
print("Number of digits : % d" % (countDigit(n)))
|
C#
using System;
class GFG {
static int countDigit(long n)
{
int count = 0;
while (n != 0)
{
n = n / 10;
++count;
}
return count;
}
public static void Main()
{
long n = 345289467;
Console.WriteLine("Number of"
+ " digits : " + countDigit(n));
}
}
|
PHP
<?php
function countDigit($n)
{
$count = 0;
while ($n != 0)
{
$n = round($n / 10);
++$count;
}
return $count;
}
$n = 345289467;
echo "Number of digits : "
. countDigit($n);
?>
|
Output
Number of digits : 9
Recursive Solution:
C++
#include <bits/stdc++.h>
using namespace std;
int countDigit(long long n)
{
if (n == 0)
return 0;
return 1 + countDigit(n / 10);
}
int main(void)
{
long long n = 345289467;
cout << "Number of digits :" << countDigit(n);
return 0;
}
|
C
#include <stdio.h>
int countDigit(long long n)
{
if (n == 0)
return 0;
return 1 + countDigit(n / 10);
}
int main(void)
{
long long n = 345289467;
printf("Number of digits : %d", countDigit(n));
return 0;
}
|
Java
import java.util.*;
class GFG {
static int countDigit(long n)
{
if (n == 0)
return 0;
return 1 + countDigit(n / 10);
}
public static void main(String[] args)
{
long n = 345289467;
System.out.print("Number of digits : "
+ countDigit(n));
}
}
|
Python3
def countDigit(n):
if n == 0:
return 0
return 1 + countDigit(n // 10)
n = 345289467
print("Number of digits : % d" % (countDigit(n)))
|
C#
using System;
class GFG {
static int countDigit(long n)
{
if (n == 0)
return 0;
return 1 + countDigit(n / 10);
}
public static void Main()
{
long n = 345289467;
Console.WriteLine("Number of "
+ "digits : "
+ countDigit(n));
}
}
|
PHP
<?php
function countDigit($n)
{
if ($n == 0)
return 0;
return 1 + countDigit((int)($n / 10));
}
$n = 345289467;
print ("Number of digits : " .
(countDigit($n)));
?>
|
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++
#include <bits/stdc++.h>
using namespace std;
int countDigit(long long n) {
return floor(log10(n) + 1);
}
int main(void)
{
long long n = 345289467;
cout << "Number of digits : "
<< countDigit(n);
return 0;
}
|
C
#include <math.h>
#include <stdio.h>
int countDigit(long long n) {
return floor(log10(n) + 1);
}
int main(void)
{
long long n = 345289467;
printf("Number of digits : %d", countDigit(n));
return 0;
}
|
Java
import java.util.*;
class GFG {
static int countDigit(long n)
{
return (int)Math.floor(Math.log10(n) + 1);
}
public static void main(String[] args)
{
long n = 345289467;
System.out.print("Number of digits : "
+ countDigit(n));
}
}
|
Python3
import math
def countDigit(n):
return math.floor(math.log10(n)+1)
n = 345289467
print("Number of digits : % d" % (countDigit(n)))
|
C#
using System;
class GFG {
static int countDigit(long n)
{
return (int)Math.Floor(Math.Log10(n) + 1);
}
public static void Main()
{
long n = 345289467;
Console.WriteLine("Number of digits : "
+ countDigit(n));
}
}
|
PHP
<?php
function countDigit($n)
{
return floor(log10($n)+1);
}
$n = 345289467;
echo "Number of digits : ",
countDigit($n);
?>
|
Output
Number of digits : 9
Method 4:
We can convert the number into a string and then find the length of the string to get the number of digits in the original number.
C++
#include <bits/stdc++.h>
using namespace std;
void count_digits(int n)
{
string num = to_string(n);
cout << num.size() << endl;
}
int main()
{
int n = 345;
count_digits(n);
return 0;
}
|
Java
import java.util.*;
public class GFG {
static void count_digits(int n)
{
String num = Integer.toString(n);
System.out.println(+num.length());
}
public static void main(String args[])
{
int n = 345;
count_digits(n);
}
}
|
Python3
def count_digits(n):
n = str(n)
return len(n)
n = 456533457776
print(count_digits(n))
|
C#
using System;
using System.Collections.Generic;
class GFG {
static void count_digits(int n)
{
string num = Convert.ToString(n);
Console.WriteLine(+num.Length);
}
public static void Main(string[] args)
{
int n = 345;
count_digits(n);
}
}
|
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