Hi,
On Mon, Jul 25, 2011 at 8:41 PM, Ian Lance Taylor <iant@xxxxxxxxxx> wrote:
> Arnaud Lacombe <lacombar@xxxxxxxxx> writes:
>
>> gcc will only emits the warning at -Os. It seems to me that the
>> resulting code clearly ends-up testing an uninitialized value, ie.
>> assuming the following test-case:
>>
>> extern void *e(void);
>> extern void *f(void);
>> extern void g(void);
>>
>> void fn(void)
>> {
>> void *b, *a;
>>
>> a = e();
>> if (a != 0)
>> b = f();
>> if (a != 0 && b != 0)
>> g();
>> }
>>
>> ...
>>
>> It seems gcc transforms the conditional from:
>>
>> if (a != NULL && b != NULL) ...
>>
>> to
>>
>> if (b != NULL && a != NULL) ...
>>
>> In which case the warning is fully valid. I'm not sure what's the C
>> standard guarantee in term of conditional test order. gcc 4.7.0 has
>> the same behavior.
>
> Not quite. C guarantees that && is executed in order. In this case gcc
> is generating
>
> a = e();
> if (a != NULL)
> b = f();
> if (a != NULL & b != NULL)
> g();
>
> Note the change from && to & in the last conditional.
I've got some trouble linking the final optimized tree with the code
generated. That is, the final tree (generated by -fdump-tree-all) is:
foo ()
{
void * b;
void * a;
_Bool D.1993;
_Bool D.1992;
_Bool D.1991;
<bb 2>:
a_2 = e ();
if (a_2 != 0B)
goto <bb 3>;
else
goto <bb 4>;
<bb 3>:
b_4 = f ();
<bb 4>:
# b_1 = PHI <b_3(D)(2), b_4(3)>
D.1991_5 = a_2 != 0B;
D.1992_6 = b_1 != 0B;
D.1993_7 = D.1992_6 & D.1991_5;
if (D.1993_7 != 0)
goto <bb 5>;
else
goto <bb 6>;
<bb 5>:
g (); [tail call]
<bb 6>:
return;
}
but the code generated seem to test %esi (`b', potentially
uninitialized) before %ebx (`a'). Am I still missing something ?
Thanks,
- Arnaud
