On Dec 8, 2003, Steve Ellcey <[EMAIL PROTECTED]> wrote:
> I'm curious. I thought I knew shell scripting and the basics of
> automake but I don't know what 'fnord' is
set fnord [expansion]; shift
is a common idiom to ensure that the variable expansion doesn't start
with say -e, causing set to do something other than setting the
argument list.
>> From: Alexandre Oliva <[EMAIL PROTECTED]>
>>
>> LIBTOOL_BEGIN_COMPILE_CC = set fnord \
>>
>> LIBTOOL_END_COMPILE_CC = ; shift 1; \
>> { test -d $$dir"/$(libtool_libdir) || \
>> $(mkdir_p) "$$dir"/$(libtool_libdir); } && \
>> rm -f "$$lofile"T "$$lofile" "$$dir/$$ofile" \
>> "$$dir/$(libtool_libdir)/$$ofile" || : ; \
>> $(LIBTOOL_COMPILE_CC_PIC) -o "$$dir/$(libtool_libdir)/$$ofile" \
>> $${1+"$$@"} && \
>> $(LIBTOOL_COMPILE_CC_NONPIC) -o "$$dir/$$ofile" $${1+"$$@"} && \
>> { echo pic_object=$(LIBTOOL_PIC_OBJECT); \
>> echo non_pic_object=$(LIBTOOL_NONPIC_OBJECT); } > "$$lofile"T && \
>> mv "$$lofile"T "$$lofile"
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--
Alexandre Oliva http://www.ic.unicamp.br/~oliva/
Red Hat Compiler Engineer [EMAIL PROTECTED], gcc.gnu.org}
Free Software Evangelist [EMAIL PROTECTED], gnu.org}
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