difference between int64_t on unix and linux

dand · 02-08-2005, 05:51 AM

I have found a bug in our code.

The line is:

int64_t iBase = (int64_t) (d1 / d2);

where d1 and d2 are doubles.

if d1 = 3000 and d2 = 0.01,

for Unix, iBase is calculated as 299999

but for Linux, iBase is calculated as 300000 (which is correct!)

Does anyone know why this is happening or where I can find the definition for int64_t on Linux?

Cheers, Dan.

btmiller · 02-08-2005, 08:54 PM

int64_t is guaranteed to be a 64 bit signed integer if you compiler follows the C99 standard. The exact mechanics of floating point rounding are probably more dependant on your hardware than on your OS. Can you post some more detail about the Unix platform this was running on (there hasn't been one single Unix since about the mid '70s), in particular the hardware.

Also, this may be better off in the programming forum (ask a mod to move it).

dand · 02-09-2005, 04:34 AM

Solaris 5.8

arch = sun4
uname = SunOS

Is that enough?

Cheers, Dan.

dand · 02-09-2005, 04:39 AM

btmiller, I'm afraid I put that the wrong way around!

It calculates it as 300000 on Unix which is correct and 299999 on Linux which is incorrect. It is a problem with Linux (RHAS3)

Sorry, Dan.