//program to count number of digits present in a number.
#include<stdio.h>
#include<conio.h>
#include<conio.h>
void main()
{
int n,rem,count=0;
printf("enter a number for 'n'\n");
scanf("%d",&n);
{
int n,rem,count=0;
printf("enter a number for 'n'\n");
scanf("%d",&n);
while(n!=0)
{
{
rem=n%10;
count++;
n=n/10;
}
printf("total digits =%d",count);
getch();
}
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logics in mind:-
>first we enter a number(for 'n').
->If we have number 123 then it has 3 digits and they are 3, 2 and 1 or 1,2 and 3. It means, first we have to get 3 then 2 and then 1 and we have to count them.
->For this,
->we have to divide by 10 to get last digit as remainder
->we also use 'count' to start counting digits.
-> to get second digit, we get first 12 and for this , we use 123/10.It is done to get integer part only.
->We repeat this until the value reaches 0. We use loop for this as shown above in the code.
->at last we display total number of digits stored in variable 'count'.
