How to printf long long

Question

I'm doing a program that aproximate PI and i'm trying to use long long, but it isn't working. Here is the code

#include<stdio.h>
#include<math.h>
typedef long long num;
main(){
    num pi;
    pi=0;
    num e, n;
    scanf("%d", &n);
    for(e=0; 1;e++){
      pi += ((pow((-1.0),e))/(2.0*e+1.0));
      if(e%n==0)
        printf("%15lld -> %1.16lld\n",e, 4*pi);
      //printf("%lld\n",4*pi);
    }
}

Answer 1

• Your scanf() statement needs to use %lld too.
• Your loop does not have a terminating condition.

• There are far too many parentheses and far too few spaces in the expression

  pi += pow(-1.0, e) / (2.0*e + 1.0);
  

• You add one on the first iteration of the loop, and thereafter zero to the value of 'pi'; this does not change the value much.
• You should use an explicit return type of int for main().
• On the whole, it is best to specify int main(void) when it ignores its arguments, though that is less of a categorical statement than the rest.
• I dislike the explicit licence granted in C99 to omit the return from the end of main() and don't use it myself; I write return 0; to be explicit.

I think the whole algorithm is dubious when written using long long; the data type probably should be more like long double (with %Lf for the scanf() format, and maybe %19.16Lf for the printf() formats.

Answer 2

First of all, %d is for a int

So %1.16lld makes no sense, because %d is an integer

That typedef you do, is also unnecessary, use the type straight ahead, makes a much more readable code.

What you want to use is the type double, for calculating pi and then using %f or %1.16f.

Answer 3

%lld is the standard C99 way, but that doesn't work on the compiler that I'm using (mingw32-gcc v4.6.0). The way to do it on this compiler is: %I64d

So try this:

if(e%n==0)printf("%15I64d -> %1.16I64d\n",e, 4*pi);

and

scanf("%I64d", &n);

The only way I know of for doing this in a completely portable way is to use the defines in <inttypes.h>.

In your case, it would look like this:

scanf("%"SCNd64"", &n);
//...    
if(e%n==0)printf("%15"PRId64" -> %1.16"PRId64"\n",e, 4*pi);

It really is very ugly... but at least it is portable.

Answer 4

    double pi = 2 * acos(0.0);
    int n;
    scanf("%d",&n); //precision with which you want the value of pi
    printf("%.*lf\n",n,pi);