C Program to find the number of digits in an integer
Finding the number of digits in an integer in C
Today, we will be learning how to find the number of digits in an integer in C . An integer is made up of a group of digits, i.e; it is a combination of digits from 0-9
Here we will use loops along with an arithmetic operator.This program takes an integer from the user and calculates the number of digits. For example: If the user enters 6589, the output of the program will be 4.
Method Discussed :
- Method 1 : Using loop
- Method 2 : Using formulae.
Let’s discuss above two methods in brief,
Method 1 :
- Take a variable count = 0, to store the count of digits in the given number.
- Run a while loop till n>2
- Inside the loop increment the value of count by 1
- Set n=n/10
- After complete iteration print the value of count.
Method 1 : Code in C
Run
#include <stdio.h>
#include <math.h>
int main(){
int n = 20901;
int count = 0;
while(n>0){
count++;
n = n/10;
}
printf("No. of digits = %d", count);
}
Output
No. of digits = 5
Method 2 :
In this method we will simple use the mathematical formulae,
- No. of digits : floor(log10(n))+1
Method 2 : Code in C
Run
#include<stdio.h>
#include<math.h>
int main(){
int n = 20901;
int x = floor(log10(n))+1;
printf("No. of digits = %d", x);
}
Output
No. of digits = 5
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*SIMPLE FORM IN C
#include
void main()
{
int n,rem=0,count=1;
printf(“Enter the input”);
scanf(“%d”,&n);
while(n>0)
{
rem=n%10;
n=n/10;
rem=count++;
}
printf(“Number of digit in a number is: %d”,rem);
}