Re: make is setting exit code to 0 when gcc command fails

[Paul Smith]

On Tue, 2015-02-10 at 10:12 +0000, David Aldrich wrote:
> define make-depend
>     @echo $(CXX) -c $4 $1 -o $2; \
>     $(CXX) -c -MMD -MP -MF $3.$$$$ $4 $1 -o $2; \
>     sed 's,\($*\)\.o[ :]*,$(notdir $2) $3 : ,g' < $3.$$$$ > $3; \
>     rm -f $3.$$$$
> endef

This is exactly what I and others have mentioned.

Here, you run the compile command as the second command in this script.
That exits with an error code.  But, you ignore that error (don't do
anything about it) then you run sed, then you run rm.

So, the exit code of your script is the exit code of the last command
you ran, which is "rm".  So the only way this command will ever fail is
if the final "rm" fails, which pretty much never happens.

This is bad.  The simplest way to fix it is to replace the ";" with
"&&", so that subsequent parts of the script don't run if an earlier
part fails:

  define make-depend
      @echo $(CXX) -c $4 $1 -o $2; \
      $(CXX) -c -MMD -MP -MF $3.$$$$ $4 $1 -o $2 \
        && sed 's,\($*\)\.o[ :]*,$(notdir $2) $3 : ,g' < $3.$$$$ > $3 \
        && rm -f $3.$$$$
  endef

There are other options, like saving the exit code of the compiler and
then using it explicitly in "exit" at the end after "rm".