linux-kernel - Re: [PATCH] btrfs: Use immediate assignment when referencing cc-option

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Date:   Wed, 17 Mar 2021 09:39:09 +0100
From:   David Sterba <dsterba@...e.cz>
To:     Victor Erminpour <victor.erminpour@...cle.com>
Cc:     clm@...com, josef@...icpanda.com, dsterba@...e.com,
        linux-btrfs@...r.kernel.org, linux-kernel@...r.kernel.org
Subject: Re: [PATCH] btrfs: Use immediate assignment when referencing
 cc-option

On Tue, Mar 16, 2021 at 03:46:10PM -0700, Victor Erminpour wrote:
> Calling cc-option will use KBUILD_CFLAGS, which when lazy setting
> subdir-ccflags-y produces the following build error:
> 
> scripts/Makefile.lib:10: *** Recursive variable `KBUILD_CFLAGS' \
> 	references itself (eventually).  Stop.
> 
> Use := assignment to subdir-ccflags-y when referencing cc-option.
> This causes make to also evaluate += immediately, cc-option
> calls are done right away and we don't end up with KBUILD_CFLAGS
> referencing itself.
> 
> Signed-off-by: Victor Erminpour <victor.erminpour@...cle.com>
> ---
>  fs/btrfs/Makefile | 8 ++++----
>  1 file changed, 4 insertions(+), 4 deletions(-)
> 
> diff --git a/fs/btrfs/Makefile b/fs/btrfs/Makefile
> index b634c42115ea..3dba1336fa95 100644
> --- a/fs/btrfs/Makefile
> +++ b/fs/btrfs/Makefile
> @@ -7,10 +7,10 @@ subdir-ccflags-y += -Wmissing-format-attribute
>  subdir-ccflags-y += -Wmissing-prototypes
>  subdir-ccflags-y += -Wold-style-definition
>  subdir-ccflags-y += -Wmissing-include-dirs
> -subdir-ccflags-y += $(call cc-option, -Wunused-but-set-variable)
> -subdir-ccflags-y += $(call cc-option, -Wunused-const-variable)
> -subdir-ccflags-y += $(call cc-option, -Wpacked-not-aligned)
> -subdir-ccflags-y += $(call cc-option, -Wstringop-truncation)
> +subdir-ccflags-y := $(call cc-option, -Wunused-but-set-variable)
> +subdir-ccflags-y := $(call cc-option, -Wunused-const-variable)
> +subdir-ccflags-y := $(call cc-option, -Wpacked-not-aligned)
> +subdir-ccflags-y := $(call cc-option, -Wstringop-truncation)

But this overwrites all the previously accumulated values from += so
effectively there's only the last one, no? That's not what we want.