Makefile: remove unused @@PERLLIBDIR@@ substitution variable

Junio noticed that this variable is not quoted correctly when it is
passed to sed.  As a shell-quoted string, it should be inside
single-quotes like $(perllibdir_relative_SQ), not outside them like
$INSTLIBDIR.

In fact, this substitution variable is not used.  Simplify by removing
it.

Reported-by: Junio C Hamano <gitster@pobox.com>
Signed-off-by: Jonathan Nieder <jrnieder@gmail.com>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
This commit is contained in:
Jonathan Nieder 2018-04-23 16:24:22 -07:00 коммит произвёл Junio C Hamano
Родитель 86e254584b
Коммит 90df2173f2
1 изменённых файлов: 0 добавлений и 1 удалений

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@ -2109,7 +2109,6 @@ GIT-PERL-HEADER: $(PERL_HEADER_TEMPLATE) GIT-PERL-DEFINES Makefile
INSTLIBDIR="$$INSTLIBDIR$${INSTLIBDIR_EXTRA:+:$$INSTLIBDIR_EXTRA}" && \
sed -e 's=@@PATHSEP@@=$(pathsep)=g' \
-e 's=@@INSTLIBDIR@@='$$INSTLIBDIR'=g' \
-e 's=@@PERLLIBDIR@@='$(perllibdir_SQ)'=g' \
-e 's=@@PERLLIBDIR_REL@@=$(perllibdir_relative_SQ)=g' \
-e 's=@@GITEXECDIR_REL@@=$(gitexecdir_relative_SQ)=g' \
-e 's=@@LOCALEDIR_REL@@=$(localedir_relative_SQ)=g' \