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| Date: | Mon, 10 Apr 2000 17:00:42 +0200 (IST) |
| From: | Eli Zaretskii <eliz AT is DOT elta DOT co DOT il> |
| X-Sender: | eliz AT is |
| To: | "Mark E." <snowball3 AT bigfoot DOT com> |
| cc: | djgpp-workers AT delorie DOT com |
| Subject: | Re: Porting problems with Sh-utils (beta) |
| In-Reply-To: | <38F1A726.23834.332F7B@localhost> |
| Message-ID: | <Pine.SUN.3.91.1000410165821.24866E-100000@is> |
| MIME-Version: | 1.0 |
| Reply-To: | djgpp-workers AT delorie DOT com |
| Errors-To: | nobody AT delorie DOT com |
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On Mon, 10 Apr 2000, Mark E. wrote:
> Add:
> echo "${1+$@}"
>
> run 'make' and you'll see what's going on. The line continutation shown below
> tricks Bash into treating 'echo `echo` script executed' as an argument
> belonging to the previous command instead of treating it like a new command.
I understood that just from reading the command, but I still don't grasp
why does this cause the $source variable to lose its value. What am I
missing?
> I'd recommend changing the makefile from:
> >
> > all:
> > source='There is no bug in BASH!' \
> > $(SHELL) ./script \
> > echo `echo`script executed
>
> to
>
> >
> > all:
> > source='There is no bug in BASH!' \
> > $(SHELL) ./script
> > echo `echo`script executed
I think the Makefile actually _wants_ this to be a single line. Is
anything wrong with that? I understand that it works on Unix unaltered.
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