Re: [PATCH v4 1/4] Makefile: generate Perl header from template file

From: Johannes Schindelin <johannes.schindelin@gmx.de>
Date: Dec 1, 2017, 17:13 UTC
Message-ID: <alpine.DEB.2.21.1.1712011812250.98586@virtualbox>
In-Reply-To: <3c897f4c-0cb5-ca44-f144-598a623c1dc3@kdbg.org>

Hi Hannes,

On Fri, 1 Dec 2017, Johannes Sixt wrote:

Show 16 quoted lines

> Am 29.11.2017 um 16:56 schrieb Dan Jacques:
> > @@ -1989,6 +1986,15 @@ GIT-PERL-DEFINES: FORCE
> >        	echo "$$FLAGS" >$@; \
> >        fi
> >   +GIT-PERL-HEADER: $(PERL_HEADER_TEMPLATE) GIT-PERL-DEFINES perl/perl.mak
> > Makefile
> > +	$(QUIET_GEN)$(RM) $@ && \
> > +	INSTLIBDIR=`MAKEFLAGS= $(MAKE) -C perl -s --no-print-directory
> > instlibdir` && \
> > +	INSTLIBDIR_EXTRA='$(PERLLIB_EXTRA_SQ)' && \
> > +	INSTLIBDIR="$$INSTLIBDIR$${INSTLIBDIR_EXTRA:+:$$INSTLIBDIR_EXTRA}" &&
> > \
> > +	sed -e 's=@@PATHSEP@@=$(pathsep)=g' \
> 
> This doesn't work, unfortunately. When $(pathsep) is ';', we get an incomplete
> sed expression because ';' is also a command separator in the sed language.

Funny, I tried this also with ';' as pathsep, and it worked in the Git for Windows SDK...

Ciao, Dscho