💻 while & for Loop Examples in C++23 min read

Armstrong number

An Armstrong number is a number that is equal to the sum of the digits that rise to the total number of digits of the power. Some Armstrong numbers are: 0, 1, 2, 3, 153, 370, 407, 1634, 8208, etc.

Know more about Armstrong number – Wikipedia

Logic

After taking input from the user, we have to calculate the total number of digits in the given number. Here we are using log10 method of math library to achieve this. After doing that we have to run one loop until the value of user given number “num” does not reach 0 or becomes less than zero.

Now inside the loop each time we have to find the last digit of user given number and then calculate the power of that number with total no of digits in the “num”, round off the result and store it in one temporary variable “sum” once the loop is finished we have to check if the value of variable sum is equivalent to user given number, if yes, then it is an Armstrong number.

Palindrome

A palindrome is anything that reads the same backward and forward. In other words after reversing any number, word or phrase if it remains the same as original then it is a palindrome.

Logic

To check whether a number is a palindrome number or not we have to reverse it first. After reversing it we compare the reversed number with the original number. If both the numbers are similar then it means the number we have is a palindrome number.

Prime number

A prime number is an integer which is greater than 1 whose only factors are 1 and itself. A factor is an integer that can be divided evenly into another number.

Logic

The logic to check a number is prime or not is really simple. We only need to check if the given number is completely divisible by any other smaller number or not, but not by 1 and itself.

After taking input from the user, we have to run one loop up to half of the given number. Now the reason for running one loop up to half of the given number is because we know, a number can only be divisible by a number which is lesser than its half.

Inside the loop we have to check if the given number “num” is completely divisible by current number “i” or not, If yes then set the value of the variable i to num and terminate the loop here, if no then continue the loop.

Once the loop is completed or terminated control comes out of the loop, here we have to check if the value of the variable i is equivalent to the given number or not. If yes then it is not a Prime number otherwise it is a prime number.

Output

Count number of digits in a given integer

To count the number of digits in a number we have to divide that number by 10 until it becomes 0 or less than 0. If you divide any number by 10 and store the result in an integer then it strips the last digit. So we have to apply the same logic here also.

Logic

After taking the input from the user, run one while loop that runs until the user given number becomes 0. Inside the loop for each iteration increase the counter variable by one and divide the user given number by 10 and store it in the same variable again.

Notice here we are using /= operator that performs division and stores the result in the same variable again.

Once the loop is over we have to print counter variable “count” containing the total number of digits.

Product of digits in a number

This program is closely similar to this one: Count number of digits in a given integer. The only difference here is instead of counting the total number of digits we are multiplying each digit by another one until the user given number becomes 0 or less than 0.

Logic

First of all, we are declaring one variable product with value 1, we are going to use this variable to store the product of all digits, now after taking input from the user, inside the while loop, we have to extract each digit by performing modulo operation.

The modulo operation returns the remainder after dividing a number by another number. If we perform modulo operation with 10 on any number it will return the last most digit, we have to multiply and store the result into the variable product.

After that remove the last most digit from the number and perform the same again until the number becomes 0 or less than 0. Once the loop is completed simple print the variable product containing the product of all digit.

Output

Sum of digits in a number

This program is much similar to this one: Find product of digits in a number. The only difference is instead of multiply we have to perform addition here.

Logic

First of all, we are declaring one variable sum with value 0, we are going to use this variable to store the sum of all digits, now after taking input from the user, inside the while loop, we have to extract each digit by performing modulo operation.

The modulo operation returns the remainder after dividing a number by another number. If we perform modulo operation with 10 on any number it will return the last most digit, we have to add it into variable sum and store the result again in variable sum.

After that remove the last most digit from the number and perform the same again until the number becomes 0 or less than 0. Once the loop is completed simply print the variable sum containing the sum of all digit.

Factorial

A factorial is a product of an integer and all the integers below it. In other words, factorial (!) Is the product of all integers from 1 to n.

For example:

If N = 5, then 5! is 5 x 4 x 3 x 2 x 1 = 120

Logic

First of all, we declare one temporary variable fact with value 1, now we are going to use this variable to store factorial.

After taking input from the user, we need to start one loop from 1 to N, and inside the loop we multiply current number “i” with variable fact and then store the result again in fact variable until the loop terminates. Once the loop is over we have to print the fact variable containing factorial of a given number (N).

HCF

The greatest common divisor of two or more numerical values is called the highest common factor (HCF). In other words, the largest number that can completely divide two or more numbers is the highest common factor.

Logic

Let declare one temporary variable hcf by 1, we are going to use this variable to store HCF.

Here we are calculating HCF of two user given numbers, after taking both the numbers (num1, num2) from the user, we have to find out which number is smaller, we store the minimum one in the variable min.

Now we start one loop from 1 to min (Minimum between both numbers), and inside the loop, we have to check if both the numbers are completely divisible by current number “i” or not. If yes then we update temporary variable hcf by current number “i”, otherwise continue with the loop.

Once the loop is over simply print the variable hcf containing the highest common factor of num1 and num2.

n this program, we are going to see how we can reverse any user given number. Reversing any word, phrase or number is pretty simple if we treat it as a string data. We can simply iterate through each character in reverse order and then append it to one temporary variable. But here we are only dealing with numbers and for numbers, there is a better approach.

Logic

First of all, we have to declare one temporary variable reverse with value 0, we will use this variable to store the reverse of the number.

After taking input from the user, we have to assign the input to one temporary variable temp, Now to reverse a number we have to extract each digit of that number one by one. To extract digits we can use % (modulo) operator. If we perform modulo operation with 10 on any number it will return the last most digit.

Now after extracting the digit, we have to append it to our temporary variable reverse, but we cannot directly append the digit on it. To append the digit, first, we need to multiply 10 on existing reversed value and then perform addition with the extracted digit.

At the end remove the last most digit and repeat the same again until the value of temp becomes 0 or less than 0.

Once the loop is finished simply print the variable reverse.

In this program, we are going to see how we can print the multiplication table of any user given input number.

Logic

Printing multiplication table of any number is quite simple and an easy task. After taking input (num) from the user we have to start one loop from 1 to 10, and then inside the loop simply multiply num by current number “i” and print it.

Output