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Re: Incorrect alias expansion within command substitution

From: Chet Ramey
Subject: Re: Incorrect alias expansion within command substitution
Date: Sat, 5 Feb 2022 15:46:06 -0500
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On 2/5/22 3:44 PM, Alex fxmbsw7 Ratchev wrote:
On Sat, Feb 5, 2022 at 9:39 PM Alex fxmbsw7 Ratchev <fxmbsw7@gmail.com> wrote:

On Sat, Feb 5, 2022 at 7:55 PM Chet Ramey <chet.ramey@case.edu> wrote:

On 2/4/22 6:17 PM, Alex fxmbsw7 Ratchev wrote:
what about this viewing point
aliases can start, $(('s, but not end... this is unlogic

Well, I don't know about `unlogic' but there's an understanding deficit,
for sure.

alias -- \
p='printf %s\\n ' \
assign='assign=$(( ' begin='$(( ' \

for data in "1 + 2"
alias -- data="$d "
p begin data ))

this works and results 3

OK, let's go through it. I'll gloss over some non-essential details and
simplify others. I'll skip over the bulk of the for loop parsing and just
cover the simple command where alias expansion takes place.

Since `p' is in a command position, it gets alias expanded to
`printf %s\\n '. The lexical analysis restarts, the two words get scanned,
and we start a simple command. The expansion ends with a space, so the next
token (`begin') undergoes alias expansion, gets expanded to `$(( ', and the
lex restarts. We read `$((' and know we're reading the start of a word that
begins with an arithmetic expansion. We keep reading as if we are reading a
double-quoted string, looking for `))', since that's what you do while
reading an arithmetic expansion.

so the alias expansion fails at the current command is no alias
anymore, so alias parsing rules dont apply anymore, .. no ... hmm .. ?
would make much sense then and i have nothing else to say..

maybe ask about an shopt feature to enable straight alias expansion
when the aliases are following ..

How would that, whatever it is, help you here?

``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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