How to printf long long

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);
    }
}

karadoc

%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.

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.

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.

    // acos(0.0) will return value of pi/2, inverse of cos(0) is pi/2 
    double pi = 2 * acos(0.0);
    int n; // upto 6 digit
    scanf("%d",&n); //precision with which you want the value of pi
    printf("%.*lf\n",n,pi); // * will get replaced by n which is the required precision

来源：https://stackoverflow.com/questions/6400180/how-to-printf-long-long

标签

long-integer