[Buildroot] Which order: $(MAKE) $(TARGET_CONFIGURE_OPTS) or $(TARGET_CONFIGURE_OPTS) $(MAKE)

From: Thomas Petazzoni <thomas.petazzoni@free-electrons.com>
To: buildroot@busybox.net
Subject: [Buildroot] Which order: $(MAKE) $(TARGET_CONFIGURE_OPTS) or $(TARGET_CONFIGURE_OPTS) $(MAKE)
Date: Wed, 29 Mar 2017 21:35:52 +0200	[thread overview]
Message-ID: <20170329213552.6455f568@free-electrons.com> (raw)
In-Reply-To: <CAAXf6LUHH5Lo=LSSfjGRQ=onT4Bm_WfLAFc0RDjfcT_ZmADmbA@mail.gmail.com>

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