[Buildroot] Which order: $(MAKE) $(TARGET_CONFIGURE_OPTS) or $(TARGET_CONFIGURE_OPTS) $(MAKE)
Thomas Petazzoni
thomas.petazzoni at free-electrons.com
Wed Mar 29 19:35:52 UTC 2017
Hello,
On Wed, 29 Mar 2017 19:58:18 +0200, Thomas De Schampheleire wrote:
> The order matters because it has impact on the precedence rules of make.
> In the case where $(MAKE) is present first, the variables are passed
> as make variables, which have precedence over variables explicitly
> defined in the makefiles.
> However, if $(MAKE) is _after_ the variable list, then the variables
> are treated as environment variables. In make, environment variables
> are directly readable as make variables, but an explicit assignment to
> such variable overwrites the value passed via the environment, unless
> the assignment is done conditionally, e.g. with ?= instead of = / :=.
>
> Given this, if you want to make sure that the value you give is taken
> into account, the forms:
>
> $(MAKE) $(TARGET_CONFIGURE_OPTS)
> $(MAKE) $(HOST_CONFIGURE_OPTS)
Sadly, doing this breaks a number of packages. Many packages that use
hand-written Makefiles do:
CFLAGS += -I../include
for example. So if you pass CFLAGS="$(TARGET_CFLAGS)" as a make
variable (i.e after $(MAKE)), then it overrides the CFLAGS of the
package Makefile, and it no longer builds.
It would work if those packages were doing:
override CFLAGS += -I../include
but they often don't do this.
Recent example:
https://git.buildroot.org/buildroot/commit/?id=f4dc73568b08bd96aa659c5ef29226349dee05de
Best regards,
Thomas
--
Thomas Petazzoni, CTO, Free Electrons
Embedded Linux and Kernel engineering
http://free-electrons.com
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