- C Programming Examples
- C Programming Examples
- C Print Hello World
- C Get Input from User
- C Print Integer
- C Add Two Numbers
- C Check Even or Odd
- C Check Prime or Not
- C Check Alphabet or Not
- C Check Vowel or Not
- C Check Leap Year or Not
- C Check Reverse equal Original
- C Add Subtract Multiply Divide
- C Make Calculator
- C Add Digits of Number
- C Calculate Average, Percentage
- C Calculate Arithmetic Mean
- C Calculate Student Grade
- C Print Table of Number
- C Print Prime Numbers
- C Add n Numbers
- C Interchange Numbers
- C Reverse a Number
- C Swap Two Numbers
- C Count Positive Negative Zero
- C Find Largest of Two Numbers
- C Find Largest of Three Numbers
- C Find Factorial of Number
- C Find LCM & HCF
- C Find LCM of n Numbers
- C Find HCF of n Numbers
- C Area & Perimeter of Square
- C Area & Perimeter of Rectangle
- C Area & Circumference of Circle
- C Convert Fahrenheit to Celsius
- C Convert Celsius to Fahrenheit
- C Print ASCII Value
- C Print Fibonacci Series
- C Check Palindrome or Not
- C Check Armstrong or Not
- C Generate Armstrong Numbers
- C Find nCr and nPr
- C Convert Decimal to Binary
- C Convert Decimal to Octal
- C Convert Decimal to Hexadecimal
- C Convert Binary to Decimal
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- C Convert Binary to Hexadecimal
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- C Convert Octal to Binary
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- C Add Two Numbers using Pointer
- C Find Largest Element in Array
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- C Insert Element in Array
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- C Merge Two Arrays
- C Bubble Sort
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- C Add Two Matrices
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- C Transpose a Matrix
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- C Find Length of String
- C Compare Two String
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- C Find Frequency of Character
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- C Remove Spaces from String
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- C Uppercase to Lowercase
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- C Swap Two Strings
- C Check Anagram or Not
- C Generate Random Numbers
- C Read a File
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- C Merge Two Files
- C List Files in Directory
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- C Print Date
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- C Inches to Centimetres
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- C Count Even Odd
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In this tutorial, we will learn about how to create a program in C that will find and print sum of squares of any two number and also will learn to find and print sum of squares of digits of any given number by user at run-time.

Let's first create a program in C that will ask from user to enter any two number to find and print sum of square of the given two numbers.

// ----codescracker.com---- #include<stdio.h> #include<conio.h> int main() { int num1, num2, sqr1, sqr2, sum; printf("Enter any two numbers: "); scanf("%d%d", &num1, &num2); sqr1 = num1*num1; sqr2 = num2*num2; sum = sqr1 + sqr2; printf("Sum of their square = %d", sum); getch(); return 0; }

As the program was written under **Code::Blocks** IDE, therefore after successful build and run, here is the
sample run:

Now supply any two number say **4** and **5** and press `ENTER`

key to see the result that will
be the sum of square of given two number. If you will square the two number say **4** and **5**, then you will
get **16** and **25**. After summing it up, you will get **16+25** equals **41** as result. Here is the
second snapshot of the sample run:

Here are some of the main steps used in above program:

- Receive any two number as input from user (at run-time) in any two variables
say
**num1**and**num2** - Calculate the square of both the number one by one and initialized its square to any two variables say
**sqr1**and**sqr2** - That is
**num1*num1**gets initialized to**sqr1**and**num2*num2**gets initialized to**sqr2** - Now calculate sum of two variable say
**sqr1**(square of first number) and**sqr2**(square of second number) - That is
**sqr1+sqr2**gets initialized to a variable say**sum**(holds summation value of square of the two given number) - Print the value of
**sum**at last of the program - For example, if user enters 4 and 5 as input
- Therefore,
**num1**holds 4 and**num2**holds 5 - Now
**num1*num1**or**4*4**or**16**gets initialized to**sqr1** - And
**num2*num2**or**5*5**or**25**gets initialized to**sqr2** - And
**sqr1+sqr2**or**16+25**or**41**gets initialized to**sum** - Print the value of
**sum**which is 41 as output

Now let's create a program that will ask from user to enter any number to find and print the sum of squares of its
digit. That is if user enters 342 as input, then the program will find the sum of square of its digit, that is
**(3*3) + (4*4) + (2*2)** or **9+16+4** or **29** will be the final result. Let's take a look at the
program:

// ----codescracker.com---- #include<stdio.h> #include<conio.h> int main() { int num, rem, sum=0; printf("Enter any number: "); scanf("%d", &num); while(num>0) { rem = num%10; sum = sum + rem*rem; num = num/10; } printf("\nSum of Square of Digits = %d", sum); getch(); return 0; }

Here is the final snapshot of the sample run:

Here are some of the main steps used in above program:

- Receive any number as input
- Never forgot to take a variable say
**sum**(holds sum of square of digits) with**0**as its initial value - That is declare and initialize 0 to
**sum**at start of the program - After gettting input from user in a variable say
**num**, follow the steps given below - Create a while loop, runs until the value of
**num**is greater than 0 - Inside the loop, calculate the digit of that number using modulous operator (%)
- Using modulous operator or remainder operator, we will get the last digit of number
- Find the square of that digit and sum it up with
**sum**variable and then initialize it to**sum**variable - Then divide the value (num) with 10 to skip or delete the last digit
- For example, if user enters 2346 as input
- Therefore, at first run of the
**while**loop, as the value of**num**(2346) is greater than 0, therefore program flow goes inside the loop, and**num%10**or**2346%10**or**6**gets initialized to**rem**. Then**sum + rem*rem**or**0 + 6*6**or**36**gets initialized to**sum**. And at last,**num/10**or**2346/10**or**234**gets initialized to**num** - And then program flow goes inside the
**while**loop's condition statement, and as the value**234**is again greater than 0, therefore again the condition of**while**loop evaluates to be true, therefore program flow goes inside the loop, and do the same steps as told above - That is,
**num%10**or**234%10**or**4**gets initialized to**rem**. And**sum + rem*rem**or**36 + 4*4**or**36 + 16**or**52**gets initialized to**sum**. And at last again**num/10**or**234/10**or**23**gets initialized to**num** - Again at third run of the
**while**loop, as the condition statement of the loop, that is**num>0**or**23>0**evaluates to be true, therefore program flow again goes inside the**while**loop and follow the same procedure until the value of**num**becomes 0 - At last run of the
**while**loop, the variable**sum**holds its value as**6*6 + 4*4 + 3*3 + 2*2**or**36 + 16 + 9 + 4**or**65**as its final value, therefore 65 is the final output which is the result of sum of square of digits of given number that is 2346