Q.2 write a program in c language using a function method to calculate the following series and give the result;
X= n^1*n^2*n^3*…………* n^k
Where n&k are any number given by the user
pls… help me

void PrintArray(int *pnArray, int nLength)
{
for (int iii=0; iii < nLength; iii++)
cout << pnArray[iii] << endl;
}

Shouldn't the 'pnArray[iii]' part be dereferenced (ie; '*pnArray[iii]')to get the values IN the pointer array? Or does it not work like that with pointer arrays? Clearly there's something I'm not getting!
Any help, well appreciated.
Thanks for all the great tutorials. You have a great way of explaining concepts step by step!

 SetToSix(nTvalue)
instead of
SetToSix(pPtr);
 

However, still i am getting error while compiling my code. I am enclosing a clean code.

#include "stdafx.h"
#include 

		using namespace std;

		int Func_Ref_Ptr(int *&pTmpr);

        int Tvalue=6;
		int nValue=9;

		int main()
		{

			cout<<"Start of our function:"<<endl;

            //intialize the reference to the pointer to a variable
			int *&pPoint = &Tvalue; 

			cout<<Func_Ref_Ptr(pPoint)<<endl;

		 return 0;

		}

		int Func_Ref_Ptr(int *&pTmpr)
		   {

			   cout<<"The value is :"<<endl;  

		     return *pTmpr;
		   }

Please reply me and guide me what i am doing wrong.

I am having a problem compiling reference to a pointer example code. I am following your code example, however i am getting error while initializing reference to a pointer in main.

 int * &pPtr = &nTvalue; 

         //function call
         SetToSix(nTvalue);

I am getting following errors on VS 2008

error C2440: 'initializing' : cannot convert from 'int *' to 'int *&'
 error C2664: 'SetToSix' : cannot convert parameter 1 from 'int *' to 'int *&'
 error C2065: 'nSix' : undeclared identifier

 int nLength= sizeof(pnArray)/sizeof(int); 

would be nice to make the function PrintArray more elegant as it would only require one parameter:

 void PrintArray(int *pnArray); // why not??