Re: Keep MLton from being kicked out of Debian?

I'm curious, what is the reason for position indep. code by default in
Debian now?

On Thu, Nov 10, 2016 at 3:00 PM, Florian Weimer <[email protected]> wrote:
> * Matthew Fluet:
>
>> And, this is where I haven't figured out what Debian is trying to
>> achieve.  Is the PIC/PIE default gcc just during package building (so
>> that executables within .deb packages are PIE), or is it also the
>> behavior of an installed Debian system, so that a normal user invoking
>> gcc will get the PIC/PIE default?
>
> It's the latter.  GCC says:
>
> Configured with: […] --enable-default-pie […]
>
> I have verified that just running gcc with no flags produces a PIE.
>
>> If the behavior of an installed Debian system is not to have a
>> PIC/PIE default gcc, then you will end up with the awkward situation
>> where MLton will (unnecessarily) generate PIC code, although I don't
>> think it a problem if PIC object is linked with non-PIC objects into
>> a non-PIE executable.
>
> No, it's going to work just fine.  It's slightly slower than
> necessary, but especially on amd64, the difference is minimal.
>
> So I think it would be fine to generate PIC code unconditionally on
> Debian.
>
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