• Carlos (26/26) May 02 2013 I have this code :
• Simen Kjaeraas (5/31) May 02 2013 Both 5 and 9 in the second example are integers (int). When you divide
• bearophile (6/9) May 02 2013 Yes, smarter languages (like Pascal..., but also Python, Ada,
• =?UTF-8?B?TWFydGluIERyYcWhYXI=?= (12/38) May 02 2013 Hi Carlos,

"Carlos" <checoimg gmail.com> writes:

I have this code :

import std.stdio;
import std.c.stdlib;
void main()
{
int fahr;
write("F\tC\n");
for (fahr = 0; fahr <= 300; fahr = fahr + 20)
write(fahr, "\t", (5.0/9.0)*(fahr-32), "\n");
write("Done!\n");
exit (0);
}
  Which works. but if I change the "5.0" for "5" I get cero on the 
celsius side.

import std.stdio;
import std.c.stdlib;
void main()
{
int fahr;
write("F\tC\n");
for (fahr = 0; fahr <= 300; fahr = fahr + 20)
write(fahr, "\t", (5/9)*(fahr-32), "\n");
write("Done!\n");
exit (0);
}

So why is this ?

"Simen Kjaeraas" <simen.kjaras gmail.com> writes:

On 2013-05-02, 20:14, Carlos wrote:

 I have this code :

 import std.stdio;
 import std.c.stdlib;
 void main()
 {
 int fahr;
 write("F\tC\n");
 for (fahr = 0; fahr <= 300; fahr = fahr + 20)
 write(fahr, "\t", (5.0/9.0)*(fahr-32), "\n");
 write("Done!\n");
 exit (0);
 }
   Which works. but if I change the "5.0" for "5" I get cero on the  
 celsius side.

 import std.stdio;
 import std.c.stdlib;
 void main()
 {
 int fahr;
 write("F\tC\n");
 for (fahr = 0; fahr <= 300; fahr = fahr + 20)
 write(fahr, "\t", (5/9)*(fahr-32), "\n");
 write("Done!\n");
 exit (0);
 }

 So why is this ?

Both 5 and 9 in the second example are integers (int). When you divide
one int by another, the result is an int, and hence (5/9) is 0.

-- 
Simen

"bearophile" <bearophileHUGS lycos.com> writes:

Simen Kjaeraas:

 Both 5 and 9 in the second example are integers (int). When you 
 divide
 one int by another, the result is an int, and hence (5/9) is 0.

Yes, smarter languages (like Pascal..., but also Python, Ada, 
etc) have two different division operators to avoid such silly C 
semantics, that sometimes causes bugs.

Bye,
bearophile

=?UTF-8?B?TWFydGluIERyYcWhYXI=?= <drasar ics.muni.cz> writes:

Dne 2.5.2013 20:14, Carlos napsal(a):
 I have this code :

 import std.stdio;
 import std.c.stdlib;
 void main()
 {
 int fahr;
 write("F\tC\n");
 for (fahr = 0; fahr <= 300; fahr = fahr + 20)
 write(fahr, "\t", (5.0/9.0)*(fahr-32), "\n");
 write("Done!\n");
 exit (0);
 }
   Which works. but if I change the "5.0" for "5" I get cero on the
 celsius side.

 import std.stdio;
 import std.c.stdlib;
 void main()
 {
 int fahr;
 write("F\tC\n");
 for (fahr = 0; fahr <= 300; fahr = fahr + 20)
 write(fahr, "\t", (5/9)*(fahr-32), "\n");
 write("Done!\n");
 exit (0);
 }

 So why is this ?

Hi Carlos,

the second code performs integral division which very much behave like 
floating-point division, but the fractional part is chopped off.

  5/9 ~ 0.556 => 0
10/9 ~ 1.111 => 1

If you want precise (i.e. floating point) results, you have to have at 
least one float or double in your equation.

This would work:

write(fahr, "\t", (5.0/9)*(fahr-32), "\n");

Regards,
Martin