Question about default optimizations

[Jonathan Wakely]

On 23 November 2015 at 13:02, staticx <mohamedboussaa3@gmail.com> wrote:
> OK. You are right, my question was not so clear. I provide an example to
> better explain :
>
> -giving a random sequence (among On options controllable through compiler
> switches): -foption1 -fno-option2 -foption3
> -I call this sequence sample_sequence
>
> -I note by FNO-O1: the list of optimizations of O1 that are controllable
> through compiler switches. I turn off all of them.
> FNO-O2 = -fno-O1Option1 -fno-O1Option2 -fno-O1Option3 ... -fno-O1OptionN
>
> -I note by FNO-O2: the list of optimizations of O1 that are controllable
> through compiler switches. I turn off all of them.
> FNO-O2 = -fno-O2Option1 -fno-O2Option2 -fno-O2Option3 ... -fno-O2OptionN
>
> TEST 1:
>
> -O1 + FNO-O1 (disable O1 optimizations that are controllable through
> compiler switches) + sample_sequence* IS IT EQUAL TO* -O2 + FNO-O2 (disable
> O2 optimizations that are controllable through compiler switches) +
> sample_sequence
>
> I would like to know of default optimizations in O1 = O2. The same question
> for O3, OS and Ofast. Does Os for example introduce more default
> optimizations than O1 (of course by considering that all optimizations that
> are controllable through compiler switches, are disabled )
>
> Hope that it is more clear.


Yes, thanks.

I don't know what answers you will find, but the method looks reasonable to me.

If you want complete confirmation of your results I suspect you'll
need to look at the GCC source to see what happens at each
optimization level.