Can i get a copy of the application.
On Mar 30, 2010 3:08 PM, "gerel" <[email protected]> wrote:
Hi,
I've tried to reproduce the bug there named. I couldn't.
However, I believe there is some issue with signs.
I made some tests myself regarding the number sign that should be returned.
The
mod operator native implementation works like this:
INPUT
##
#include <stdio.h>
int main (){
printf ("5 % 3: %d\n", 5%3);
printf ("5 % -3: %d\n", 5%-3);
printf ("-5 % 3: %d\n", -5%3);
printf ("-5 % -3: %d\n\n", -5%-3);
printf ("3 % 5: %d\n", 3%5);
printf ("3 % -5: %d\n", 3%-5);
printf ("-3 % 5: %d\n", -3%5);
printf ("-3 % -5: %d\n", -3%-5);
}
##
OUTPUT:
anon...@castle:~$ ./a.out
5 % 3: 2
5 % -3: 2
-5 % 3: -2
-5 % -3: -2
3 % 5: 3
3 % -5: 3
-3 % 5: -3
-3 % -5: -3
##
That means that the result sign is _only_ decided on the dividend sign. In
the
current code for pdf_i64_mod() (built-in implementation), It says:
##
[...]
/*Now check the signs fo divisor and dividend*/
if (pdf_i64_cmp(divisor,zero) < 0)
{
result_sign = -1;
pdf_i64_abs(&divisor,divisor, p_status);
}
if (pdf_i64_cmp(dividend,zero) < 0)
{
pdf_i64_abs(÷nd,dividend, p_status);
if (result_sign == -1)
{
result_sign = -1;
}
else
{
result_sign = -1;
}
}
[...]
###
If the variable named "result_sign" is deciding the returned number sign.
I believe it should be something like this instead,
##
[...]
/*Now check the signs fo divisor and dividend*/
if (pdf_i64_cmp(divisor,zero) < 0)
{
pdf_i64_abs(&divisor,divisor, p_status);
}
if (pdf_i64_cmp(dividend,zero) < 0)
{
pdf_i64_abs(÷nd,dividend, p_status);
result_sign = -1;
}
[...]
###
I may be wrong, I didn't read the whole function code, because I'm not
familiar
with the Knuth method.
Despite that, I tried the above proposed modification and no new test fails.
It
seems to work fine or do nothing. :-)
Any comments ?
cheers,
-gerel
