Makefile: use $(INSTALL_LIB) for modules not $(INSTALL_PROG)

We have $(INSTALL_LIB) which is the same as $(INSTALL_PROG) but
uses correct permissions.  Loadable objects (modules) are like
shared libraries, not like programs.

Signed-off-by: Michael Tokarev <mjt@tls.msk.ru>
Cc: Fam Zheng <famz@redhat.com>
Cc: Paolo Bonzini <pbonzini@redhat.com>
Signed-off-by: Paolo Bonzini <pbonzini@redhat.com>

diff --git a/Makefile b/Makefile
index 23ca4446daa3cbbb2301df1955f88fcff41eac02..12ea46424f0af5caadc231e233b58d13b56ad1a6 100644 (file)
--- a/Makefile
+++ b/Makefile
@@ -381,7 +381,7 @@ ifneq ($(CONFIG_MODULES),)
        $(INSTALL_DIR) "$(DESTDIR)$(qemu_moddir)"
        for s in $(modules-m:.mo=$(DSOSUF)); do \
                t="$(DESTDIR)$(qemu_moddir)/$$(echo $$s | tr / -)"; \
-               $(INSTALL_PROG) $$s "$$t"; \
+               $(INSTALL_LIB) $$s "$$t"; \
                test -z "$(STRIP)" || $(STRIP) "$$t"; \
        done
 endif