On 11/11/06, Charles R Harris <[EMAIL PROTECTED]> wrote:
This made me curious. Here is the code
int test(double a, double b)
{
if (a > b)
return 0;
return 1;
}
Here is the relevant part of the assembly code when compiled with no optimizations
fucompp
fnstsw %ax
sahf
ja .L4
jmp .L2
.L4:
movl $0, -20(%ebp)
jmp .L5
.L2:
movl $1, -20(%ebp)
.L5:
movl -20(%ebp), %eax
leave
ret
Which jumps to the right place on a > b (ja)
Here is the relevant part of the assembly code when compiled with -O3
fucompp
fnstsw %ax
popl %ebp
sahf
setbe %al
movzbl %al, %eax
ret
Which sets the value of the return to the logical value of a <= b (setbe), which won't work right with NaNs. Maybe the compiler needs another flag to deal with the possibility of NaN's because the generated code is actually incorrect. Or maybe I just discovered a compiler bug. But boy, that compiler is awesome clever. Those optimizers are getting better all the time.
Chuck
On 11/11/06, Tim Hochberg <[EMAIL PROTECTED] > wrote:Robert Kern wrote:
<snip>My preference would be to raise an error / warning when there is a nan
in the array. Technically, there is no minimum value when a nan is
present. I believe that this would be feasible be swapping the compare
from 'a < b' to '!(a >= b)'. This should return NaN if any NaNs are
present and I suspect the extra '!' will have minimal performance impact
but it would have to be tested. Then a warning or error could be issued
on the way out depending on the erstate. Arguably returning NaN is more
correct than returning the smallest non NaN anyway.
No telling what compiler optimizations might do with '!(a >= b)' if they assume that '!(a >= b)' == 'a < b'. For instance,
if !(a >= b)
do something;
else
do otherwise;
This made me curious. Here is the code
int test(double a, double b)
{
if (a > b)
return 0;
return 1;
}
Here is the relevant part of the assembly code when compiled with no optimizations
fucompp
fnstsw %ax
sahf
ja .L4
jmp .L2
.L4:
movl $0, -20(%ebp)
jmp .L5
.L2:
movl $1, -20(%ebp)
.L5:
movl -20(%ebp), %eax
leave
ret
Which jumps to the right place on a > b (ja)
Here is the relevant part of the assembly code when compiled with -O3
fucompp
fnstsw %ax
popl %ebp
sahf
setbe %al
movzbl %al, %eax
ret
Which sets the value of the return to the logical value of a <= b (setbe), which won't work right with NaNs. Maybe the compiler needs another flag to deal with the possibility of NaN's because the generated code is actually incorrect. Or maybe I just discovered a compiler bug. But boy, that compiler is awesome clever. Those optimizers are getting better all the time.
Chuck
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