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Re: No word splitting for assignment-like expressions in compound assignment

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Alexey Izbyshev <izbyshev@ispras.ru> writes:
> I have a question about the following behavior:
>
> $ Z='a b'
> $ A=(X=$Z)
> $ declare -p A
> declare -a A=([0]="X=a b")
> $ A=(X$Z)
> $ declare -p A
> declare -a A=([0]="Xa" [1]="b")
>
> I find it surprising that no word splitting is performed in the first 
> compound assignment.

> * Brace expansion is performed for "A=(X=a{x,y}b)" by all bash versions 
> mentioned above (which is inconsistent with normal variable assignment).
> * Globbing for "A=(X=a?b)" is performed by bash 3.1.17, but not by other 
> versions.

Interesting.  The documentation for 4.2.53(1) says this about parameter
assignments generally, with no special rules for compound assignments:

       All
       values undergo tilde expansion, parameter and variable expansion,  com-
       mand  substitution, arithmetic expansion, and quote removal (see EXPAN-
       SION below).  ...  Word  splitting  is  not
       performed,  with the exception of "$@" as explained below under Special
       Parameters.  Pathname expansion is not  performed.

So it seems like the word splitting in "A=(X$Z)" is incorrect.  So is
pathname expansion in that context.  Oddly, brace expansion is not
mentioned.

Dale

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Re: No word splitting for assignment-like expressions in compound assignment worley@alum.mit.edu (Dale R. Worley) - 2020-07-27 01:31 -0400

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