I came across this construct while reviewing some older unix shell scripts, what does it mean, and why is is used?
${1+"$@"}
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What did the documentation say about it? – Lightness Races in Orbit Jul 14 '15 at 22:10
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3[Shell Parameter Expansion](http://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion) Important sentence that is easy to miss: "Omitting the colon results in a test only for a parameter that is unset." – Etan Reisner Jul 14 '15 at 22:12
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Thanks! I found it hard to search google (or even stackoverflow) with such constructs. – xxfelixxx Jul 14 '15 at 22:16
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I found this explanation from unix haters handbook, page 152 of text (page 190 of the pdf). http://web.mit.edu/~simsong/www/ugh.pdf
It’s the way to exactly reproduce the command line arguments in the /bin/sh family of shells shell script.
It says, “If there is at least one argument ( ${1+ ), then substitute in all the arguments ( “$@” ) preserving all the spaces, etc. within each argument.
If we used only “$@” then that would substitute to “” (a null argument) if there were no invocation arguments, but we want no arguments reproduced in that case, not “”.
Why not “$*” etc.? From a sh(1) man page:
Inside a pair of double quote marks (“”), parameter and command substitution occurs and the shell quotes the results to avoid blank interpretation and file name generation. If $* is within a pair of double quotes, the positional parameters are substituted and quoted, separated by quoted spaces (“$1 $2 …”); however, if $@ is within a pair of double quotes, the positional parameters are substituted and quoted, separated by unquoted spaces (“$1” “$2” …).
I think ${1+“$@”} is portable all the way back to “Version 7 Unix.”
Wow! All the way back to Version 7.
xxfelixxx
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It seems you don't get the "null argument" in modern sh:
/bin/sh -c '/bin/sh -c "echo \$#" -- "$@"' -- – Boyd Stephen Smith Jr. May 03 '22 at 16:470
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Google should be your best friend! This is a similar questions asked almost 2 years ago.
