Hi,
On Tue, 14 Oct 2014, Jamie Iles wrote:
> int foo(void)
> {
> if (getreturn() != 0)
> return -1;
>
> return 0;
> }
So if getreturn() returns zero it can simply reuse that return value ...
> but at -O1 I get
>
> 10: fb ff ff 40 call 0 <foo>
> 10: R_OLDLAND_PC24 getreturn-0x4
> 14: 01 00 00 3c mov $r1, 0x0
> 18: 01 01 00 0a sub $r1, $r1, $r0
> 1c: 00 01 00 2a or $r0, $r1, $r0
> 20: 00 f0 01 38 asr $r0, $r0, 0x1f
... and if I'm interpreting the mnemonics correctly that is what seems to
happen here. If $r0 is zero, then:
mov $r1, 0x0 $r1 = 0
sub $r1, $r1, $r0 $r1 = 0 - 0 = 0
or $r0, $r1, $r0 $r0 = 0 | 0 = 0
asr $r0, $r0, 0x1f $r0 = 0 >> 31 = 0
Voila, zero is correctly returned. Any non-zero value will be transformed
into -1 (if positive the high bit will be set due to subtraction, if
negative the high bit will be set due to the 'or', and the shift
replicates the high bit into the lower ones, yielding -1).
Ciao,
Michael.
