C++ Program to Find and Print the Product of the Digits of a Number

This article provides some programs in C++ that find and print the product of the digits of a number entered by the user at run-time. The program is created in the following ways:

• Find and print the product of the digits of a number using the while loop
• Find and print the product of the digits of a number using the for loop

And at last, one extra program is also included that finds the product of the only non-zero digit of a given number.

Using a while loop, calculate the product of a number's digits

The question is: write a C++ program that receives a number from the user and prints the product or multiplication of its digit. The program given below is its answer:

#include<iostream>

using namespace std;
int main()
{
   int num, rem, prod=1;
   cout<<"Enter a Number: ";
   cin>>num;
   while(num>0)
   {
      rem = num%10;
      prod = prod*rem;
      num = num/10;
   }
   cout<<"\nProduct of all digits of given number is: "<<prod;
   cout<<endl;
   return 0;
}

The initial output produced by the above C++ program on finding and printing the product of all digits of a given number is given in the following snapshot:

Now supply the input, say 1293 as a number, and press the ENTER key to print the product of all four digits as shown in the snapshot given here:

The dry run of the following block of code from the above program:

while(num>0)
{
   rem = num%10;
   prod = prod*rem;
   num = num/10;
}

with prod=1 and num=1293, goes like this:

• The condition num>0 or 1293>0 is satisfied.
• Therefore, program flow goes inside the loop, and all three statements get executed one by one.
• That is, num%10 or 1293%10 or 3 gets initialized to rem, prod*rem or 1*3 or 3 gets initialized to prod, and num/10 or 1293/10 or 129 gets initialized to num.
• Now the condition of the loop gets evaluated again with the new value of num.
• That is, the condition num>0 or 129>0 evaluates to true once more.
• Therefore, program flow again goes inside the loop.
• This process continues until the condition is evaluated as false.
• In this way, the product of the digits of a number gets calculated.

Find the Product of a Number's Digits Using a for Loop

This program performs the same function as the previous one, but it was written in a for loop rather than a while loop.

#include<iostream>

using namespace std;
int main()
{
   int num, rem, prod;
   cout<<"Enter a Number: ";
   cin>>num;
   for(prod=1; num>0; num=num/10)
   {
      rem = num%10;
      prod = prod*rem;
   }
   cout<<"\nProduct of all digits of given number is: "<<prod;
   cout<<endl;
   return 0;
}

In the above program, the following two statements:

rem = num%10;
prod = prod*rem;

can also be replaced with the single statement given below:

prod = prod*(num%10);

Find the product of the digits (except 0) of a number

Since both the programs given above give 0 as output if any digit of the given number is 0, let's modify and create another program that finds the product of only one non-zero digit of the given number:

#include<iostream>

using namespace std;
int main()
{
   int num, rem, prod;
   cout<<"Enter a Number: ";
   cin>>num;
   for(prod=1; num>0; num=num/10)
   {
      rem = num%10;
      if(rem>0)
         prod = prod*rem;
   }
   cout<<"\nProduct of all digits of given number is: "<<prod;
   cout<<endl;
   return 0;
}

This program produces the same output as the previous program. But the previous program printed 0 if any digit of the given number was equal to 0, whereas this program does not print 0 if any digit of the given number is 0. Rather, this only multiplies non-zero digits and prints the multiplication result.

Modified Version of the Previous Program

This program is created in a way to provide a good user interface by showing the ongoing multiplication of digits like 3*2*4.

#include<iostream>

using namespace std;
int main()
{
   int num, rem, prod=1;
   cout<<"Enter a Number: ";
   cin>>num;
   if(num>0)
      cout<<endl;
   else
   {
      cout<<"\nThe number must be greater than 0!\n";
      return 0;
   }
   while(num>0)
   {
      rem = num%10;
      num /= 10;
      if(rem>0)
      {
         prod = prod*rem;
         if(num>0)
            cout<<rem<<"*";
         else
            cout<<rem;
      }
   }
   cout<<" = "<<prod;
   cout<<endl;
   return 0;
}

This program produces the following output when the user enters a number, say 12032, as input:

You're seeing the digit in the form 2*3*2*1 (in reverse order). because while finding the remainder using rem = num%10, it gives the last digit. But you can reverse the number first, then do the task.

To do it, put the reverse of a number code in the very first if's body and initialize the reversed result to the number num itself, then proceed. Come on, do it yourself for practice.

C++ Quiz