Help a beginner with looping

[nucleoponey]

Hello, I'm trying to come up with a little calculator for a specific equation that has to do with sigma. SQRT(r^2-x^2), where r is a constant (input as radius) and x=(1, 2,... r).
For example, if r=3,
⌊SQRT(3^2-1^2)⌋+⌊SQRT(3^2-2^2)⌋+⌊SQRT(3...

 #include <iostream>

using namespace std;

#include <cmath>

int main()
{
    cout<<"Radius: ";
    int r;
    cin>>r;
    for(int x=1;x<=r; x++){
    int rad[r];
    rad[r]=x;

    int p;


p=sqrt(r^2-(rad[(r)])^(2));

    cout<<p<<endl;

    }
}

What is the issue with the code?

[JockX]

What is the issue with the code?

#include <iostream>
using namespace std;
#include <cmath>

int main()
{
    cout<<"Radius: ";
    int r;
    cin>>r;
    for(int x=1;x<=r; x++){
        int rad[r]; // <== standard C++ does not allow 
                    // non constant size while declaring array

        rad[r]=x;   // <== in EACH iteration assign a new 
                    // value to element outside the array

        int p;      // declare p as int but assign double value
                    // so the result is rounded down 

        p=sqrt(r^2-(rad[(r)])^(2)); // Again rad[r] is outside the array
        cout<<p<<endl;  
	}
}

[Chervil]

r^2 does not mean r²
Use r*r instead
^ is the bitwise exclusive or operator
http://www.cplusplus.com/doc/boolean/

[nucleoponey]

Thanks for the replies.

int rad[r]; // <== standard C++ does not allow
// non constant size while declaring array

rad[r]=x; // <== in EACH iteration assign a new
// value to element outside the array

JockX, thanks for the information, but how can I work around it?

[Mats]

JockX, thanks for the information, but how can I work around it?

http://www.cplusplus.com/reference/vector/vector/

[Chervil]

I'm trying to ignore all the code, and look only at the question.

As I understand things, the program needs to find the sum of the terms of the series sqrt(r² - x²) where r is an integer, and x takes the values from 1 to r-1. (When x is equal to r, the term becomes zero).

e.g. when x = 3,
sqrt(3² - 1²) + sqrt(3² - 2²)

Now for correct calculation, floating point numbers must be used (type double) but both r and x are integers. Though I could have just used double throughout, I decided to explicitly make use of the appropriate types.

#include <iostream>
#include <iomanip>

using namespace std;

#include <cmath>

int main()
{
    cout << "Radius: ";
    int r;
    cin >> r;
    
    double total = 0;
    cout << fixed;
    for (int x=1; x<r; x++)
    {
        double xd = x;
        double rd = r;
        double term = sqrt(rd*rd - xd*xd);  // calculate current term of series
        total += term;                      // add it to the total        
        cout << setw(3) << x << "  term = " << setw(8) << term 
             << "  total: " << setw(8) << total << endl;
    }
}

Output:

Radius: 7
  1  term = 6.928203  total: 6.928203
  2  term = 6.708204  total: 13.636407
  3  term = 6.324555  total: 19.960962
  4  term = 5.744563  total: 25.705525
  5  term = 4.898979  total: 30.604505
  6  term = 3.605551  total: 34.210056

[nucleoponey]

Chervil, thanks for the code. Your interpretation of my sum is correct, except that x takes values from 1 to r. But that's not very important. The code is great! But I've one more question to ask, though, how do I set the code in such a way where the value of each sqrt is rounded down before addition to the next sqrt? For example, before term1 is rounded down to 6 before addition to term2 which is also rounded down to 6, and so on.

[nucleoponey]

^
Oh wait nevermind I've figured it out. Thanks for the help!!

[Chervil]

For reference, you could use the floor() function.
http://www.cplusplus.com/reference/cmath/floor/
Or convert the value to an integer, which would simply discard the fractional part.