what's the difference between "$@" and ${1+"$@"}

cadphile has asked for the wisdom of the Perl Monks concerning the following question:

In a perl script, in the top lines, where one designates the interpreter, the perl wisdom for systems where the kernel might not do the right thing with a "#!" line, is to use the form:

#!/bin/sh -- # -*- perl -*-
eval 'exec /data/local/bin/perl -S $0 ${1+"$@"}'
  if $running_under_some_script;

So, why ${1+"$@"} instead of just "$@"? I know that "$@" expands to all the arguments quoted (e.g. "$1" "$2", etc). But the other form I don't quite see. I mean, what's the 1+ do? I've searched the for the documentation on what this extra bracketed format accomplishes, and I can't find it. And surprisingly, Google can't handle that search string.

If someone knows the answer, or an online reference to this, I'd be very happy...

cheers!
-cadphile...

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Re: what's the difference between "$@" and ${1+"$@"} by merlyn (Sage) on Oct 05, 2007 at 23:47 UTC
There was a bug in the One True Shell (/bin/sh on V7) such that a simple "$@" would fail miserably if nothing at all was passed. So we old-timers learned to write the extended version instead. Since the code that generates that #! line was no doubt written in the Perl1 days, the legacy was carried forward, although no modern shell probably still suffers the same bug. -- Randal L. Schwartz, Perl hacker	[reply]
Re^2: what's the difference between "$@" and ${1+"$@"} (misery) by tye (Sage) on Oct 06, 2007 at 01:49 UTC
such that a simple "$@" would fail miserably if nothing at all was passed "fail miserably" meant (on my systems back then, anyway) that `""` (a single empty argument) got passed instead of no arguments. It was considered more of a design quirk than a bug in my neighborhood, since it made sense for `$@` to expand to something like `foo" "bar` (if two arguments were passed) but didn't make as much sense for `$@` to make surrounding quotes, if present, just magically disappear (if no arguments passed). - tye	[reply] [d/l] [select]
Re^2: what's the difference between "$@" and ${1+"$@"} by cadphile (Beadle) on Oct 09, 2007 at 16:56 UTC
The syntax isn't quite ${1:+$foo}, and hence "this" or "that". I've never seen any documentation of the syntax `${foo+"$bar"}`. I'm just curious what the `1+` does in this context. If there are actual arguments in `"$@"`, the `1+` doesn't seem to modify the end result at all. So, if `"$@"` is empty, do we get instead the return value of the first positional parameter `$1`, which happens also to be empty? Or does the `1+` just act to clean up the return value of the empty `"$@"`.	[reply] [d/l] [select]
Re^3: what's the difference between "$@" and ${1+"$@"} by ikegami (Patriarch) on Oct 09, 2007 at 17:05 UTC
From the manual, If the colon (:) is omitted from the above expressions, the shell only checks whether parameter is set or not.	[reply]
Re^4: what's the difference between "$@" and ${1+"$@"} by cadphile (Beadle) on Oct 11, 2007 at 22:39 UTC
Re: what's the difference between "$@" and ${1+"$@"} by liverpole (Monsignor) on Oct 05, 2007 at 23:44 UTC
Hi cadphile, My /(ba)?sh/-fu is notoriously weak, so I referred to the bash manual: `3.5.3 Shell Parameter Expansion ${parameter:+word} If parameter is null or unset, nothing is substituted, otherwise the expansion of word is substituted.` [download] I take that to mean that if argument $1 (ie. ${1}) is null, nothing gets substituted, but if not, it's replaced by all arguments. In other words, perhaps it's a clean way of referring to all arguments, and not failing if none are specified? That's my best guess, anyway. s''(q.S:$/9=(T1';s;(..)(..);$..=substr+crypt($1,$2),2,3;eg;print$..$/	[reply] [d/l]