David Brown schrieb:
Georg-Johann Lay wrote:
As far as "the optimizer" of gcc is concerned, that makes no
difference. It knows exactly what register contains what value and is
aware of the place where a register "dies", i.e. the register can be
reused for whatever other stuff. Anyway, even if just one temp variabe
is used, gcc will produce a new (pseudo) register vor every result
like moves, arithmetic, etc. These pseudos may or may not end up in
the same macine register. On that level, blocks are just syntactic
sugar (if they are not used to hide visibility, e.g. like in int
tmp=0; {int tmp = 1;} )
I haven't looked at code generated for such switches (there is often so
much of it), so I admit to having guessed a little. I was thinking
especially of when you have debug information enabled - that can force
the compiler to keep variables in separate registers.
Are you really sure? As far as I know gcc produces the same code
regardless if optimization is on or not. If fact I would guess that it
is a policy that the code *must* be the same regardless what debug level
(if any) or debug format is used, and code beeing dependent on debug
level/format is worth a bug report.
To get a notion of the various machine intependant transformations,
have a glance at gcc's output with -fdump-tree-all, and for the
machine dependent it is -fdump-rtl all. They make clear that do-while,
while, for and if-goto are just flavours of same sugar.
And here was me thinking the generated source code was sometimes a bit
big to wade through... Sometime I must look at this in more detail.
Yes, of course, that example is much too complex. But for small
examples it is very interesting to track how gcc is transforming and
kneading and stiring the code again and again beyond recognition.
Georg-Johann
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