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C program to calculate the power using recursion

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This post calculates the power using recursion by the use of C programming language.

C program to calculate the power using recursion

Run Code Snippet
#include <stdio.h>
int power(int n1,int n2);
int main() 
{
    int base,a,result;
    printf("Enter base number: ");
    scanf("%d",&base);
    printf("Enter power number: ");
    scanf("%d",&a);
    result=power(base,a);
    printf("%d^%d = %d",base,a,result);
    return 0;
}
int power(int base, int a) 
{
    if(a!= 0)
        return(base*power(base,a-1));
    else
        return 1;
}

Output

Explanation

The programs get started by initializing the functions and the variables. In this program, the variables initialized to store the values are base to get the base value, a to the get the power value and result for printing the output.  We end the main function with giving return 0.

int power(int n1,int n2);
int main() 
{
    int base,a,result;
    printf("Enter base number: ");
    scanf("%d",&base);
    printf("Enter power number: ");
    scanf("%d",&a);
    result=power(base,a);
    printf("%d^%d = %d",base,a,result);
    return 0;
}

Here we call the function to calculate the power using recursion. We also use the if condition to calculate the power, where we check the condition whether it’s true or false, and then proceed further. 

int power(int base, int a) 
{
    if(a!= 0)
        return(base*power(base,a-1));
    else
        return 1;
}

After the if condition is satisfied, we get the output as desired.

C Program to Calculate Average Using Arrays

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This post shows how to calculate the average of all the values in the array using C programming language.

C Program to Calculate Average Using Arrays

Run Code Snippet
#include <stdio.h>
int main() 
{
   int array[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0};
   int sum,loop;
   float avg;
    sum =avg= 0;
    //calculation process
    for(loop=0;loop<10;loop++) 
    {
      sum=sum+array[loop];
    }
   avg=(float)sum/loop;
   //to display result
   printf("Average of array values is %.2f", avg);   
   return 0;
}

Output

Explanation

We start the program by initializing the array value and the variables, where the sum and loop operate the integer values and average holds the float value. 

	int array[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0};
    int sum,loop;
    float avg;
    sum =avg= 0;

Now we start the calculation part, where we use for condition to do the calculation and to get the average(avg) value.

	//calculation process
    for(loop=0;loop<10;loop++) 
    {
      sum=sum+array[loop];
    }
   	avg=(float)sum/loop;

After the calculation, we proceed further to get the result and we use %.2f to stop after two decimal points.

//to display result
   printf("Average of array values is %.2f", avg);   
   return 0;

We now get the desired result.

C Program to Find the Sum of Natural Numbers using Recursion

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This post helps us to find the sum of natural numbers using recursion using C programming.

C Program to Find the Sum of Natural Numbers using Recursion

#include<stdio.h>
int recusum(int s);
int main()
{
    int num;
    printf("Enter a positive number:");
    scanf("%d",&num);
    printf("sum=%d",recusum(num));
    return 0;
}
//function calling process
int recusum(int s)
{
    if(s!=0)
        return s+recusum(s-1);
    else
        return s;
}

Output

Explanation

This C program uses the recursion function, its the process of repeating items in a self-similar way, in simple words function calls itself. In this program, we initialize a function namely recusum(int s), and in the main function, we iterate a variable named num.

int recusum(int s);
int main()
{
    int num;
    printf("Enter a positive number:");
    scanf("%d",&num);
    printf("sum=%d",recusum(num));
    return 0;
}

The function is called for the calculation and to get the desired result. Using the if-else condition we calculate the recursion value of the number entered by the user.

//function calling process
int recusum(int s)
{
    if(s!=0)
        return s+recusum(s-1);
    else
        return s;
}

After the calculation part. we have now obtained the desired output.

C Program to Find G.C.D Using Recursion

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This post helps to find G.C.D using recursion by C programming language.

C Program to Find G.C.D Using Recursion

Run Code Snippet
#include<stdio.h>
int hcf(int a,int b);
int main()
{
    int c,d;
    printf("Enter the positive number:");
    scanf("%d %d", &c,&d);
    printf("G.C.D of %d and %d are:%d",c,d,hcf(c,d));
    return 0;
}
//function calling and calculation process
int hcf(int a,int b)
{
    if(b!=0)
        return hcf(a,a%b);   
    else
        return a;
}

Output

Explanation

In this C program, we find the G.C.D(greatest common divisor) value using recursion(repeats itself). The program starts with the main function where the user is allowed to enter the values for the variables initialized.

int main()
{
    int c,d;
    printf("Enter the positive number:");
    scanf("%d %d", &c,&d);
    printf("G.C.D of %d and %d are:%d",c,d,hcf(c,d));
    return 0;
}

We here use the function named HCF, where this function is called to do the particular calculation. Using the if-else condition we do the calculation and the return the value needed in the variable namely “a”

//function calling and calculation process
int hcf(int a,int b)
{
    if(b!=0)
        return hcf(a,a%b);   
    else
        return a;
}

Now after the calculation, we get our desired output.

C Program to Convert Octal Number to Decimal and vice-versa

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This post explains the conversion of octal number to decimal number and vice-versa using the C programming language.

C Program to Convert Octal Number to Decimal and vice-versa

Run Code Snippet
#include<stdio.h>
#include<math.h>
int convert(int decinum);
int main()
{
    int decinum;
    printf("Enter a decimal number: ");
    scanf("%d", &decinum);
    printf("%d in decimal = %d in octal", decinum, convert(decinum));
    return 0;
}
//Function to convert
int convert(int decinum)
{
    int octalnum = 0, i = 1;
    while (decinum != 0)
    {
        octalnum += (decinum % 8) * i;
        decinum /= 8;
        i *= 10;
    }
    return octalnum;
}

Output

Explanation

We start by initializing the function to convert and then the main function to allow the user to enter the values in the variable named decinum, where it stores the values.

int convert(int decinum);
int main()
{
    int decinum;
    printf("Enter a decimal number: ");
    scanf("%d", &decinum);
    printf("%d in decimal = %d in octal", decinum, convert(decinum));
    return 0;
}

Now we call the function to convert the numbers from octal to decimal and vice-versa.

//Function to convert
int convert(int decinum)
{
    int octalnum = 0, i = 1;
    while (decinum != 0)
    {
        octalnum += (decinum % 8) * i;
        decinum /= 8;
        i *= 10;
    }
    return octalnum;
}

we now get the desired output for the conversion.

C Program to Convert Binary Number to Octal and vice-versa

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This post helps the conversion of the binary number and octal and vice-versa using the C programming language.

C Program to Convert Binary Number to Octal and vice-versa

#include<stdio.h>
#include<math.h>
int convert(long long bin);
int main() 
{
    long long bin;
    printf("Enter a binary number: ");
    scanf("%lld", &bin);
    printf("%lld in binary = %d in octal", bin, convert(bin));
    return 0;
}
int convert(long long bin) 
{
    int octal=0,deci=0,i=0;
    //converting binary to decimal
    while (bin != 0) 
    {
        deci += (bin % 10)*pow(2, i);
        ++i;
        bin /= 10;
    }
    //converting to decimal to octal
    i = 1;
    while(deci != 0) 
    {
        octal+=(deci % 8) * i;
        deci /= 8;
        i *= 10;
    }
    return octal;
}

Output

Explanation

We start the program by initializing the function and then use the main function to get the value of variable long long bin(%lld). The value revised from the user is for the conversion of the binary number to octal and vice versa.

int convert(long long bin);
int main() 
{
    long long bin;
    printf("Enter a binary number: ");
    scanf("%lld", &bin);
    printf("%lld in binary = %d in octal", bin, convert(bin));
    return 0;
}

Now we call the function to do the conversion, the value entered by the user first gets converted from binary to decimal.

int convert(long long bin) 
{
    int octal=0,deci=0,i=0;
    //converting binary to decimal
    while (bin != 0) 
    {
        deci += (bin % 10)*pow(2, i);
        ++i;
        bin /= 10;
    }

After the value is converted as decimal, it now gets converted from decimal to octal, to complete the process of conversion of binary numbers to octal.

//converting to decimal to octal
    i = 1;
    while(deci != 0) 
    {
        octal+=(deci % 8) * i;
        deci /= 8;
        i *= 10;
    }
    return octal;
}

Now we get the converted desired output.

C Program to Convert Binary Number to Decimal and vice-versa

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This post explains the conversion of numbers from binary to decimal and vice versa, using the C programming language.

C Program to Convert Binary Number to Decimal and vice-versa

Run Code Snippet
#include<stdio.h>
#include<math.h>
int convert(long long n);
int main()
{
    long long n;
    printf("Enter a binary number:");
    scanf("%lld",&n);
    printf("\n %lld in binary=%d in decimal", n,convert(n));
    return 0;
}
int convert(long long n)
{
    int deci=0,i=0,remi;
    while(n!=0)
    {
        remi=n%10;
        n/=10;
        deci+=remi*pow(2, i);
        ++i;
    }
    return deci;
}

Output

Explanation

In this C program, we get to understand the conversion of binary numbers to decimal numbers. We initialize the function and start with the main function where we allow the user to enter the values required.

int convert(long long n);
int main()
{
    long long n;
    printf("Enter a binary number:");
    scanf("%lld",&n);
    printf("\n %lld in binary=%d in decimal", n,convert(n));
    return 0;
}

Then we call the function to do the conversion by using the while loop condition to get the desired output.

int convert(long long n)
{
    int deci=0,i=0,remi;
    while(n!=0)
    {
        remi=n%10;
        n/=10;
        deci+=remi*pow(2, i);
        ++i;
    }
    return deci;
}

We now get the desired output.

C Program to Display Armstrong Number Between Two Intervals

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This post explains how to display Armstrong number between two intervals using the C programming language.

C Program to Display Armstrong Number Between Two Intervals

#include<stdio.h>
#include<math.h>
int main()
{
    int n1,n2,i,temp,num,rem,count=0;
    printf("Enter the intervals:");
    scanf("%d %d",&n1,&n2);
    printf("Amrstrong numbers between %d and %d are:",n1,n2);
    //to store the values
    for(i=n1+1;i<n2;++i)
    {
        temp=i;
    //checking the armstrong formulation
    while (temp != 0) 
    {
         temp /= 10;
         ++count;
    }
      temp= i;
      while (temp != 0) 
      {
         rem = temp % 10;
         num += pow(rem, count);
         temp /= 10;
      }
      if ((int)num== i) 
      {
         printf("%d ", i);
      }
    // resetting the values
      count=0;
      num=0;
    }
    return 0;
}

Output

Explanation

This C program deals with Armstrong numbers, where Armstrong number is the sum of the digit’s cube is equal to the number itself. We first initialize the n1 and n2 where the user-defined values get stored using printf and scanf functions.

	int n1,n2,i,temp,num,rem,count=0;
    printf("Enter the intervals:");
    scanf("%d %d",&n1,&n2);
    printf("Amrstrong numbers between %d and %d are:",n1,n2);

Next using For loop condition we store the value in the variable named temp, and then we check the Armstrong formulation using the while condition where the temp is not equal to zero, and carry on the calculation part.

	for(i=n1+1;i<n2;++i)
    {
        temp=i;
    //checking the armstrong formulation
    while (temp != 0) 
    {
         temp /= 10;
         ++count;
    }

Again using the while loop condition we calculate the remainder(rem) and the num(result), to get the desired result.

	temp= i;
      while (temp != 0) 
      {
         rem = temp % 10;
         num += pow(rem, count);
         temp /= 10;
      }

Now using the If condition we check whether num is equal to variable “i” print the final value and then reset the values.

	if ((int)num== i) 
      {
         printf("%d ", i);
      }
    // resetting the values
      count=0;
      num=0;

We have now obtain the desired output.

C Program to Display Prime Numbers Between Two Intervals

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This post helps us to display prime numbers between two intervals using C programming.

C Program to Display Prime Numbers Between Two Intervals

Run Code Snippet
#include<stdio.h>
int main()
{
    int num1,num2,i,flag;
    printf("Enter the intervals:");
    scanf("%d %d",&num1,&num2);
    printf("Prime number between %d and %d are:",num1,num2);
    //value until num1 is not equal to num2
    while(num1<num2)
    {
        flag=0;
    //leave the numbers less than num2
    if(num1<=1)
    {
        ++num1;
        continue;
    }
    //if num1 is non-prime number
    for(i=2;i<=num1/2;++i)
    {
        if(num1%i==0)
        {
            flag=1;
            break;
        }
    }
    //for checking the prime number for the next number
    //increment num1 by 1
    if(flag==0)
    printf("%d",num1);
    ++num1;
    }
    return 0;
}

Output

Explanation

In this C program, we find prime numbers between two intervals. Here we use the flag as a variable that acts as a signal for a function or process. A flag is often binary, which contains a boolean value(true or false).

We get the values of the intervals from the user. Now we use a while loop to check whether the given num1 value is not equal to num2 value. If the num1 is not equal to num2, then the value of the flag is declared as zero so that the prime numbers are displayed between the given intervals. If the num1 value is less than two, then it gets incremented, and the process continues until the required result is obtained.

//value until num1 is not equal to num2
    while(num1<num2)
    {
        flag=0;
    //leave the numbers less than num2
    if(num1<=1)
    {
        ++num1;
        continue;
    }

Now using for loop, we declare the value for “i” as two because one is neither or nor a prime number. The value of I gets incremented until it satisfies the “I <=num1/2” condition. Using the if statement, we make sure that the num1 value is equal to zero. If yes, then the flag will be equal to 1, and the control comes out of the loop. This if statement is to check whether the values are prime numbers or non-prime numbers.

//if num1 is non-prime number

//if num1 is non-prime number
    for(i=2;i<=num1/2;++i)
    {
        if(num1%i==0)
        {
            flag=1;
            break;
        }

So we check whether the following numbers are prime numbers or not by incrementing num1 values by 1. 

	//for checking the prime number for the next 
    //increment num1 by 1
    if(flag==0)
    printf("%d",num1);
    ++num1;

If the flag =0. Then the result gets displayed.

C Program to Find the Roots of a Quadratic Equation

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This post explains how to find the roots of a Quadratic Equation using the C programming language. Let’s get started.

C Program to Find the Roots of a Quadratic Equation

#include<math.h>
#include<stdio.h>
int main()
{
    //Program for finding roots of a quadratic equation
    double R1,R2,Dis,a,b,c,realpart,imgpart;
    printf("Enter the coefficient of a,b and c:");
    scanf("%lf %lf %lf", &a,&b,&c);
    Dis=b*b-4*a*c;
    //Rules for real and different roots
    if(Dis>0)
    {
        R1=(-b+sqrt(Dis)/(2*a));
        R2=(-b+sqrt(Dis)/(2*a));
        printf("R1=%.2lf and R2=%.2lf",R1,R2);
    }
    //Checking whether the roots are real and equal
    else if(Dis==0)
    {
        R1=R2=-b/(2*a);
        printf("R1=R2=%.2lf", R1);
    }
    //Rules if the roots are not real
    else
    {
        realpart=-b/(2*a);
        imgpart=sqrt(-Dis)/(2*a);
        printf("R1=%.2lf+%.2lfi and R2=%.2lf-%.2lfi", realpart,imgpart,realpart,imgpart);
    }
    return 0;
}

Output

Explanation

First, let us get to know about quadratic equation, Quadratic Equation in algebraic form is any equation that can be reformed in standard form in which “x” is an unknown number and “a, b and c” are the known numbers, where “a is not equal to 0” and if “a=0” then it’s a linear equation and not a quadratic because there is no numeric term.

Now the main objective for us is to find whether the roots are “real and different(Dis>0)” or “real and equal(Dis=0)” or “complex and different(Dis<0)”. Where we require two roots namely R1 and R2. We need to first get the coefficient for the three variables named “a,b and c”, and finally to check whether the given two roots are not real, we use realpart and imgpart(imaginary). This is calculated with the quadratic term which is displayed below in “Dis”.

 	double R1,R2,Dis,a,b,c,realpart,imgpart;
    printf("Enter the coefficient of a,b and c:");
    scanf("%lf %lf %lf", &a,&b,&c);
    Dis=b*b-4*a*c;

The first if condition is for checking whether the roots are real and different and then the required result is stored in R1 and R2. Here we are using “%.2lf” this means that the value will be rounded off to only two decimal places, and lf means long float. 

	if(Dis>0)
    {
        R1=(-b+sqrt(Dis)/(2*a));
        R2=(-b+sqrt(Dis)/(2*a));
        printf("R1=%.2lf and R2=%.2lf",R1,R2);
    }

Now we use the if else condition, where we check the roots are real and equal, where R1 is equal to R2 and both the roots, is equal the b/(2*a).

	else if(Dis==0)
    {
        R1=R2=-b/(2*a);
        printf("R1=R2=%.2lf", R1);
    }

We use the else condition to check realpart and imgpart of the roots, in simple words whether the roots are complex and different. In this part R1 and R2 is named as realpart and imgpart respectively. Note that to attain the realpart and imgpart values we should use the formula given in the printf statement where “+” is for realpart and “-“ for imgpart.

	else
    {
        realpart=-b/(2*a);
        imgpart=sqrt(-Dis)/(2*a);
        printf("R1=%.2lf+%.2lfi and R2=%.2lf-%.2lfi", 		realpart,imgpart,realpart,imgpart);
    }

Now the C program has attained its required output.