On 15 November 2011 08:00, Avi Kivity <a...@redhat.com> wrote:
> @@ -1099,6 +1099,37 @@ for flag in $gcc_flags; do
> fi
> done
>
> +if test "$pie" = "yes" -a "$static" = "yes" ; then
> + echo "static and pie are mutually incompatible"
> + exit 1
> +fi
The -a operator to test has been marked obsolescent in
POSIX -- please don't use it in new code. (Use
if test "$pie" = yes && test "$static" = yes; then )
> +if test "$pie" = "yes" ; then
> + cat > $TMPC << EOF
> +int main(void) { return 0; }
> +EOF
> + if compile_prog "-fPIE -DPIE" "-Wl,-pie"; then
> + QEMU_CFLAGS="-fPIE -DPIE $QEMU_CFLAGS"
> + LDFLAGS="-Wl,-pie $LDFLAGS"
> + if compile_prog "-fPIE -DPIE" "-Wl,-pie -Wl,-z,relro -Wl,-z,now"; then
> + LDFLAGS="-Wl,-z,relro -Wl,-z,now $LDFLAGS"
Why does this second compile test put -fPIE -DPIE into
its local cflags and -Wl,-pie into its local ldflags
when we just put them into the global cflags/ldflags?
> + fi
> + else
> + echo "Disabling PIE due to missing toolchain support"
> + pie="no"
This means that if the user explicitly asked for PIE (with
--enable-pie") we will carry on even if we couldn't do it.
Usually for configure if the user asked for something then
not providing it is a fatal error.
-- PMM
