Answer to Question #279056 in C for teheee

Question

To give you more of a challenge with a number's digits, let's try counting how many of a certain digit is present on a given number.

Let's start coding!

Instructions:

1. Input two integer values. The first one shall accept any integer from 0-9 and the other one shall take a non-zero positive integer.
2. Using a while loop, count how many of the first integer (0-9) is present in the digits of the second inputted integer and print the result (see sample input and output for example).
3. Tip #1: You have to use your knowledge from the previous problems in looping through the digits of a number: % 10 to get the rightmost digit, while / 10 to remove the rightmost digit. Make sure to solve the previous problems first.

Answer 1

#include<stdio.h>  


int main(){
	int number,digit,count=0;    
	printf("Enter the integer and a non-zero positive integer: ");    
	scanf("%d%d",&digit,&number);    
	while(number>0){
		if (number % 10 == digit){
			count +=1;
		}
		number = number/10;
	}
	printf("%d",count); 
	


	getchar();
	getchar();
	return 0;  
}