double division error (I get -0.0000)

I'm using gcc on Linux.

Whether I do this,

Code:

---

printf(" x=%5.4f<br>",145348224945.0000/3330389019.0000);

---

or this,

Code:

---

printf(" x=%5.4f<br>",145348224945.0000L/3330389019.0000L);

---

I get -0.0000 as the result. And I don't understand why.

In the actual program, I have variables, of course.

Code:

---

int a, d;
long b, c;
long double x;

x=(long double) (a*b) / (long double) (c*d);

---

Is there any way to get around this error?

Are you sure you've got an integral value such as 145348224945?

[edit]Maybe give this a whirl.

Code:

---

x= ((long double)a*b) / ((long double)c*d);

---

Quote:

---

Originally Posted by dorakura

Is there any way to get around this error?

---

--edit stuff that is wrong--, and this code

Code:

---

#include <stdio.h>

int main ( void )
{
  printf(" x=&#37;5.4f<br>",145348224945.0000/3330389019.0000);
  return 0;
}

---

prints

Code:

---

x=43.6430<br>

---

(also on gcc on Linux). If you want to print a long double, use %Lf.

Afterthoughts:

1. MinGW users may bump into an issue with long doubles when playing along at home.
   
2. This code does not necessarily show what it is attempting to show:
   
   Quote:

   ---

   Originally Posted by tabstop

   Code:

   ---

   #include <stdio.h>

int main ( void )
{
  printf(" x=%5.4f<br>",145348224945.0000/3330389019.0000);
  return 0;
}

   ---

   prints
   

   Code:

   ---

   x=43.6430<br>

   ---

   ---

   Being constant, 145348224945.0000/3330389019.0000 might be calculated at compile time, so this code could effectively compile into
   
   Code:

   ---

   printf(" x=%5.4f<br>",43.6430171117);

   ---

3. Any problem I see is that a*b and c*d result in longs which are then converted to long doubles. Overflowing the intermediate longs before the conversions is what I suspect as the cause of the unexpected output.
   
   But with that said, I'm trying this using MinGW, so I'm hosed by the above issue. What I tried to show by moving the cast, was that explicitly converting a into a long double should implicitly convert b and the result as well; same in the denominator. By doing the math with long doubles, I would expect this issue to be handled in an expected manner.
   
   For example:
   
   Code:

   ---

   #include <stdio.h>

int main(void)
{
  int a = 15 * 11, d = 3;
  long b = 880898333, c = 1110129673;
  double x = ((double)a * b) / ((double)c * d);
  double y = (double)(a * b) / (double)(c * d);
  printf("x = %g\n", x);
  printf("y = %g\n", y);
  return 0;
}

/* my output
x = 43.643
y = 0.705659
*/

   ---

Quote:

---

Originally Posted by Dave_Sinkula

Afterthoughts:

1. MinGW users may bump into an issue with long doubles when playing along at home.
   
2. This code does not necessarily show what it is attempting to show:
   Being constant, 145348224945.0000/3330389019.0000 might be calculated at compile time, so this code could effectively compile into
   
   Code:

   ---

   printf(" x=%5.4f<br>",43.6430171117);

   ---

---

Not my code, exactly, but true; in fact if I try to initialize a variable to either of these, bad things happen:

Code:

---

temp.c:7: warning: integer constant is too large for ‘long’ type
temp.c:7: warning: overflow in implicit constant conversion
temp.c:8: warning: this decimal constant is unsigned only in ISO C90

---

So the second number would fit in an unsigned long, but long overflow is definitely happening with the first (assuming our systems are similar enough). -0.00000 is what you get when you printf a long double with %f instead of %Lf, though, is the point I was making.

Quote:

---

Originally Posted by tabstop

-0.00000 is what you get when you printf a long double with %f instead of %Lf, though, is the point I was making.

---

Ah, thanks. I couldn't see that with the MinGW/gcc.

It was a casting problem as pointed out above. I've made the following modifications and everything works now. Thank you so much for your help!

Code:

---

int a, d;
long b, c;
double x; /* long double turned out to be unnecessary */

x=((double) a * (double) b) / ((double) c * (double) d);

---