How many times a digit is appeared in a number - c

Question

Well, I wrote the code and everything is fine except one thing.
When I enter that digit number, which has to be upto 10 digits, I recieve in arr[0] various values, for example, if I enter "12345" I get 20, 1 , 1 , 1 , 1 , 1 , 0 ,0 ,0 ,0.
Which is fine from arr[1] to arr[9], but pretty odd in arr[0].
Any ideas?
#include <stdio.h>
#include <conio.h>
#include <math.h>
void main()
{
int i,j,p=0, temp,indexNum, arr[10] = { 0 }, num, level = 10, level2 = 1,maxIndex;
printf("Please enter a digit number (upto 10 digits) \n");
scanf("%d", &num);
temp = num;
while (temp > 0)
{
p++;
temp /= 10;
}
for (i = 0;i < p;i++)
{
temp = num;
while (temp > 0)
{
indexNum = num % level / level2;
arr[indexNum]++;
level *= 10;
level2 *= 10;
temp /= 10;
}
}
for (j = 0; j < 10; j++)
{
printf("%d\n", arr[j]);
}
getch();
}

Answer 1

Here is simplified version of your program:
#include <stdio.h>
#include <math.h>
int main()
{
int i = 0, j = 0, temp = 0, indexNum = 0, num = 0, level = 10;
int arr[10] = {0};
num = 7766123;
temp = num;
if(0 == temp) arr[0] = 1; // Handle 0 input this way
while (temp > 0)
{
indexNum = temp % level;
arr[indexNum]++;
temp /= 10;
}
for (j = 0; j < 10; j++)
{
printf("%d\n", arr[j]);
}
return 0;
}

Answer 2

A few hints to help you:
What does arr[10] = { 0 } actually do?
When you calculate indexNum, you are dividing integers. What happens when the modulus is a one-digit number, and level2 is greater than 1?

Answer 3

It's probably easier to read the input into a string and count digit characters. Something like this (not tested):
std::map<char, int> count;
std::string input;
std::cin >> input;
for (auto iter = input.begin(); iter != input.end(); ++iter) {
if (*iter < 0 || *iter > 9)
break;
else
++count[*iter];
}
for (auto iter = count.begin(); iter != count.end(); ++iter) {
std::cout << *iter << '\n';
}

Answer 4

You need to get rid of your first for loop. Something more like:
#include <stdio.h>
#include <math.h>
using namespace std;
int main()
{
int j;
int temp;
int indexNum;
int arr[10] = { 0 };
int num;
int level = 10;
int level2 = 1;
printf("Please enter a digit number (upto 10 digits) \n");
scanf("%d", &num);
temp = num;
while (temp > 0)
{
indexNum = num % level / level2;
arr[indexNum]++;
level *= 10;
level2 *= 10;
temp /= 10;
}
for (j = 0; j < 10; j++)
{
printf("%d\n", arr[j]);
}
return 0;
}

Answer 5

Check the program below.
void count_digits(unsigned int a, int count[])
{
unsigned int last_digit = 0;
if (a == 0) {
count[0] = 1;
}
while (a != 0)
{
last_digit = a%10;
count[last_digit]++;
a = a/10;
}
}
int main()
{
int count[10]= {0};
unsigned int num = 1122345; /* This is the input, Change it as per your need */
int i = 0;
count_digits(num, count);
for (i = 0; i < 10; i++)
{
printf ("%d: -- %d\n", i, count[i]);
}
return 0;
}