On 10 Nov 2013, at 23:05, Martin wrote:
> On 10/11/2013 21:55, Jonas Maebe wrote:
>> On 10 Nov 2013, at 22:42, Martin wrote:
>>
>>> Why is the
>>> movl %eax,-12(%ebp)
>>> generated at the start of i:=i+1
>>>
>>> Why does a statement save an outdated value to memory?
>> Because the load back from memory that came right after it got removed by a
>> peephole optimisation. Peephole optimisations are by definition local and
>> have no clue about what happens next, and hence the store remains.
>>
> Ok, i understand, but that would mean hat the save originally was intended
> for the "i := 1" statement.
The transformation that gets applied is
movl const1,(mem1)
movl (mem1),reg1
to:
movl const1,reg1
movl reg1,(mem1) }
The second instruction originally belonged to the second statement and hence
has its line number info.
> I only wonder, if despite the fact that O1 does not guarantee, if in this
> case it would be worth to consider it a bug (and intend to fix it)?
If someone wants to change it, they're free to do so. My days of working on the
peephole optimisers are over.
Jonas_______________________________________________
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