Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted. You can assume that x,p & n are integer variables and that the function will return an integer.

#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()

{

            int intUserInput, intUserInput1, intUserInput2;

            int intCompResult;

            int invert(int, int, int);

            clrscr();

            printf("\n\n\t Please insert integer to invert: ");

            scanf("%d", &intUserInput);

            printf("\n\n\t Please insert starting point to invert: ");

            scanf("%d", &intUserInput);

            printf("\n\n\t Please insert Length to invert: ");

            scanf("%d", &intUserInput2);

            intCompResult=invert(intUserInput, intUserInput1, intUserInput2);

            printf("\n\n\t Invert no. is : %d", intCompResult);

            getch();

}

int invert(int x, int p, int n)

{

            int intbinary[8];

            int i;

            int y;

            int r=0;

            for(i=0;i<8;i++)

            {

                        intbinary[i]=0;

            }

            i=0;

            y=0;

            while(x>0)

            {

                        intbinary[i]=x%2;

                        x=x/2;

                        i++;

            }

            for(i=0;i<8;i++)

            {

                        if(i==p)

                        {

                                    for(i=p;i>p-n;i--)

                                    {

                                                if(intbinary[i]==0)

                                                {

                                                            intbinary[i]=1;

                                                }

                                                else

                                                {

                                                intbinary[i]=0;

                                                }

                                    }

                                    i=i+n;

                        }

            }

            for(i=0;i<8;i++)

            {

                        r=r+(intbinary[i]*pow(2,i));

            }

            return r;

}