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Re: set -u not working as expected

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On 8/2/20 3:34 AM, Lawrence Velázquez wrote:

>>> % bash -c 'set -u; typeset -i v; v+=1; printf "<%s>\\n" "$v"'
>>> <1>
>>> % ksh -c 'set -u; typeset -i v; v+=1; printf "<%s>\\n" "$v"'
>>> <1>
>>> % zsh -c 'set -u; typeset -i v; v+=1; printf "<%s>\\n" "$v"'
>>> <1>
> 
> ...which contrasts with the behavior of let. Someone else will have
> to explain this, as I don't know what to make of it.

The integer attribute means the rhs of the assignment operator is
evaluated as an expression. It doesn't mean the assignment is performed
in an arithmetic context, as with `let' or `$((...))'. The special case,
which is done to conform to user expections, is that the result is added
to the current value, if any, instead of being appended like the usual
behavior of `+='. Note that there are no corresponding `name*=value' or
`name-=value' variable assignment operators, and that `-u' isn't in
effect for any use of `+=', whether the integer attribute is present or
not.

-- 
``The lyf so short, the craft so long to lerne.'' - Chaucer
		 ``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, UTech, CWRU    chet@case.edu    http://tiswww.cwru.edu/~chet/

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Re: set -u not working as expected Chet Ramey <chet.ramey@case.edu> - 2020-08-02 16:51 -0400

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