Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]

Groups > gnu.bash.bug > #12134

Re: ${1+"$@"} does not generate multiple words if IFS is empty

Show key headers only | View raw

On Wed, Dec 30, 2015 at 10:02:41AM +0100, Andreas Schwab wrote:
> martijn@inlv.org writes:
> 
> > 	The substitution ${1+"$@"} should resolve to "$@" if there is at
> > 	least one parameter -- i.e. one word per parameter. This works fine
> > 	if IFS contains any character or is unset. If IFS is empty, it
> > 	instead resolves to the equivalent of "$*", i.e. a single word
> > 	concatenating all the parameters without a separator. IFS should
> > 	not influence the behaviour of "$@" under any circumstances.
> 
> http://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06_05
> 
> "If the value of IFS is null, no field splitting shall be performed."

By itself that doesn't explain it.  But if you scroll up to
http://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_05_02
it says:

  @
  Expands to the positional parameters, starting from one. When the
  expansion occurs within double-quotes, and where field splitting (see
  Field Splitting) is performed, each positional parameter shall expand
  as a separate field [...]

Back to gnu.bash.bug | Previous | Next | Find similar

Thread

Re: ${1+"$@"} does not generate multiple words if IFS is empty Greg Wooledge <wooledg@eeg.ccf.org> - 2015-12-30 08:00 -0500

csiph-web