size of array pie has non-intergral type 'double' error,

Question

<double c, six, part1, part2, part3, part4,part5;
long double pi;
    cout << "enter a number: ";
    cin >> c;

    for(double i = 0; i < c; i++)
    {

           part1 = (1/(pow(16 ,(double)i)));
           part2 = 4/(((8*i))+1);
           part3 = 2/(((8*i))+4);
           part4 = 1/(((8*i))+5);
           part5 = 1/(((8*i))+6);

            pi = ((part1)*(part2 - part3 - part4 - part5));

           for(int ii = 0; ii < c; ii++)
           {
                   int pie[i] = pi;
            }



            cout << pi << "\n";>

For some reason i cant store my values in the array please help! i dont know if its initialzation errors or conversion errors or what.

Ok sorry about that i fixed it didnt notice i diddnt place a second i in the array now its giving me variable size initalization errors. any way around that?

Answer 1

I think you mean

pie[i] = pi;

instead of

int pie[i] = pi;

However you still have to declare pie somewhere. And the logic of your code is a bit confusing to me so I'm not sure what else to suggest.

But anyway you have to understand the difference between where you declare your array eg int pie[100]; and where you assign your array eg pie[i] = pi; . These are two different things.

Answer 2

 for(int ii = 0; ii < c; ii++)
 {
  pie[ii] = pi; // not pie[i],since its in the loop with iterator ii
 }

This helps?

With pie[i] the outer loop value of i is used.