dev-cpp-users

Re: [Dev-C++] calculating PI

[Alfred P. R. <al...@ya...>]

Hi Noorez,
 
Is there such a thing as an unsigned double?  IMHO, floats and doubles are always signed. 
There's also a switch for 12 byte and 16 byte long doubles, if that helps.  Try GCC --target-help.  
 
I compiled and ran your program as a new project using the File-->New-->Project...-->Console Application feature, just deleting the keyword unsigned, adding a line for clarity at the top to show that something is going on, and a precision of 15 decimal digits to the output statement.  iomanip contains the definitions to setprecision().
 
#include <cstdlib>
#include <iostream>
#include <iomanip>
using namespace std;
 
int main(int argc, char *argv[])
{
  const double numerator = 4;
  double PI = 4;  // short, signed or unsigned invalid for PI.
  bool add = false;
  cout << "Running..." << endl;
  for(double i = 3; i < 999999999; i+=2)
  {
    if( add )
    {
      PI += numerator/i;
      add = false;
    }
    else
    {
      PI -= numerator/i;
      add = true;
    }
  }
  cout << setprecision(15) << PI << endl << endl;
  system("PAUSE");
  return EXIT_SUCCESS;
}

It worked fine!

Noorez Kassam <dar...@ho...> wrote:
trying to make something to calculate PI to a large place value. The compile 
error is on the marked line

#include 
#include 
using namespace std;

int main()
{
const double numerator = 4;
unsigned double PI = 4; // short, signed or unsigned invalid for `PI'

bool add = false;

for(double i = 3; i < 999999999; i+=2)
{
if( add )
{
PI += numerator/i;
add = false;
}
else
{
PI -= numerator/i;
add = true;
}
}
cout << PI << endl << endl;
system("pause");
}

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Alfred P. Reaud
 

Re: [Dev-C++] calculating PI

[Rafael C. <cha...@un...>]

Noorez,

Just a suggestion to your algorithm in
order to not use an "if/else" statement.

 <snip>
 const double numerator = 4;
 double PI = 4;  // short, signed or unsigned invalid for PI.
 // The only modifier you can use is "long" with "double"
 // bool add = false;
 short int sign = 1; //set a variable to act as a sign
 cout << "Running..." << endl;
 for(double i = 3; i < 999999999; i+=2){
       sign *= -1; //switches the "sign" on each iteration
       PI += ((numerator/i)*sign);
 }
 </snip>

Regards,
Rafael Charnovscki

"Alfred P. Reaud" wrote:
> 
> Hi Noorez,
> 
> Is there such a thing as an unsigned double?  IMHO, floats and doubles
> ...................
> int main(int argc, char *argv[])
> {
>   const double numerator = 4;
>   double PI = 4;  // short, signed or unsigned invalid for PI.
>   bool add = false;
>   cout << "Running..." << endl;
>   for(double i = 3; i < 999999999; i+=2)
>   {
>     if( add )
>     {
>       PI += numerator/i;
>       add = false;
>     }
>     else
>     {
>       PI -= numerator/i;
>       add = true;
>     }
>   }
>   cout << setprecision(15) << PI << endl << endl;
>   system("PAUSE");
>   return EXIT_SUCCESS;
> }


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Re: [Dev-C++] calculating PI

[Szczepan H. <ru...@wp...>]

> Just a suggestion to your algorithm in
> order to not use an "if/else" statement.

<snip>

> for(double i = 3; i < 999999999; i+=2)
> {
>       sign *= -1;
>       PI += ((numerator/i)*sign);
> }

... and since an assignment expression itself has a value equal to the
value assigned, this may be rewritten as a one-liner without curly
braces:

for(double i = 3; i < 999999999; i+=2)
    PI+=((numerator/i)*(sign = -sign));

Note that "sign = -sign" might be more efficient if your compiler
cannot optimize multiplication by -1 to a single FP negate instruction
to avoid full FP multiplication.

Another optimization is to factor out the sign switching altogether.
You could increment the loop by 4 instead of 2 and add two terms in
one iteration:

for(double i = 3; i < 999999999; i+=4)
    PI += -numerator/i + numerator/(i+2);

This method reduces the number of loop iterations by two. This is an
additional benefit because each loop iteration involves a conditional
branch, and conditional branches are bad for performance on modern
CPUs.

(Note that the result might be slightly different now because of an
off-by-one issue: you might be adding an extra term in the last
iteration because the "(i+2)" is no longer checked against the loop
exit condition.)

BTW., I didn't know that PI could be calculated like this! From what I
know about harmonic series it must converge painfully slowly, hence
999999999... :)

Regards,
Szczepan Holyszewski