Use $(pie-default) with conformtest

My glibc bot showed that my conformtest changes fail the build of the
conformtest execution tests for x86_64-linux-gnu-static-pie, because
linking the newly built object with the newly built libc and the
associated options normally used for linking requires it to be built
as PIE.  Add $(pie-default) to the compiler command used so that PIE
options are used when required.

There's a case for using the whole of $(CFLAGS-.o) (which includes
$(pie-default)), but that raises questions of any impact from using
optimization flags from CFLAGS in these tests.  So for now just use
$(pie-default) as the key part of $(CFLAGS-.o) that's definitely
needed.

Tested with build-many-glibcs.py for x86_64-linux-gnu-static-pie.
This commit is contained in:
Joseph Myers 2021-09-17 19:24:14 +00:00
parent f3eef96390
commit 885762aa31

View file

@ -175,7 +175,8 @@ $(conformtest-header-tests): $(objpfx)%/conform.out: \
conformtest.py $(conformtest-headers-data)
(set -e; std_hdr=$*; std=$${std_hdr%%/*}; hdr=$${std_hdr#*/}; \
mkdir -p $(@D); \
$(PYTHON) $< --cc='$(CC)' --flags='$(conformtest-cc-flags)' \
$(PYTHON) $< --cc='$(CC) $(pie-default)' \
--flags='$(conformtest-cc-flags)' \
--ldflags='$(+link-tests-before-inputs)' \
--libs='$(+link-tests-after-inputs)' \
--run-program-prefix='$(run-program-prefix)' \