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Re: [PATCH] Replece LDFLAGS-* = $(no-pie-ldflag) with tst-*-no-pie = yes [BZ #22630]


On Tue, Dec 19, 2017 at 10:01 AM, Joseph Myers <joseph@codesourcery.com> wrote:
> On Tue, 19 Dec 2017, H.J. Lu wrote:
>
>> On Tue, Dec 19, 2017 at 9:07 AM, Joseph Myers <joseph@codesourcery.com> wrote:
>> > On Mon, 18 Dec 2017, H.J. Lu wrote:
>> >
>> >> $(no-pie-ldflag) is no longer effective since no-pie-ldflag is defined
>> >> to -no-pie only if GCC defaults to PIE.  When --enable-static-pie is
>> >> used to configure glibc build and GCC doesn't default to PIE. no-pie-ldflag
>> >> is undefined and these tests:
>> >>
>> >> elf/Makefile:LDFLAGS-tst-dlopen-aout = $(no-pie-ldflag)
>> >> elf/Makefile:LDFLAGS-tst-prelink = $(no-pie-ldflag)
>> >> elf/Makefile:LDFLAGS-tst-main1 = $(no-pie-ldflag)
>> >> gmon/Makefile:LDFLAGS-tst-gmon := $(no-pie-ldflag)
>> >>
>> >> may fail to link.  This patch replaces "-pie" with
>> >
>> > Why is --enable-static-pie causing non-static test cases to be built as
>> > PIE?  I see nothing in the NEWS or INSTALL entries to indicate such an
>> > effect.  That looks like the underlying problem here.
>> >
>>
>> To build static PIE, all .o files need to compiled with -fPIE.  Given all
>> input .o files are compiled with -fPIE when creating an executable, should
>> we generate PIE or nor no-PIE?
>
> I'd expect a normal executable, in the case of dynamic executables, since
> nothing about --enable-static-pie says it changes how non-static
> executables are built.

I can certainly make a patch to do that.  On the other hand, should
--enable-static-pie imply PIE?


-- 
H.J.


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