On 22/01/2014 00:46, Ralf Quint wrote:
On 1/21/2014 4:23 PM, Martin Frb wrote:
Sorry, but those two code blocks are NOT identical as you claim.
At least in the snippets below, GetNextInstruction is called at least
twice in the OLD snippet, not only once as your supposed NEW snippet
does. It is called once for either the ffo1 or the ffo2 case and
again, for the the ffo3 case which is executed after either ffo1 or ffo2.
Yes they change the number of invocations.
In the real code, there is at least 1 if block where GetNextInstruction
is the first condition. I did write:
- Old/ current code between 1 and 5 invocations
- New code exactly 1
GetNextInstruction is side effect free, it is safe to change the amount
of times it is called
You are however right.
If ffo1 or ffo2 changes the next instruction, then it must update the
result of GetNextInstruction too (unless it exits the block with
"continue" as some, but not all of the blocks do)
This may be a good reason (unless it is known/remembered) to check, why
some of the code that applies changes ends with a continue, and other
does not (indent or oversight?)
This has at least 2 effects:
1)
- without continue in ffo1/ffo2, it will immediately test for ffo3
- with continue in ffo1/ffo2, it will test for ffo3 in the NEXT loop
(since p(the current instruction) is kept.
2)
- with continue the result of the optimization is tested for ALL
peephole opts
- without continue it is only tested for ffo3
In case 2, the case "with continue" appears to be preferable. Unless:
- it is known that this would lead to invalid code (but then, that might
also happen if the code after round 1 occurs in in-optimized code)
- it is known that no other opt can possible apply (and this is
reviewed, each time a new one is added). But then it would do no harm to
test anyway.
NEW:
If GetNextInstruction(p, hp1) then Begin
if ffo1 then
else
if ffo2 then
;
if ffo3 then
end;
OLD:
if ffo1 and
GetNextInstruction(p, hp1) then
else
if ffo2 and
GetNextInstruction(p, hp1) then
;
if ffo3 and
GetNextInstruction(p, hp1) then
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