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Re: invisible var content to echo when not quoted - why


From: Alex fxmbsw7 Ratchev
Subject: Re: invisible var content to echo when not quoted - why
Date: Sun, 28 Nov 2021 18:44:04 +0100

yea some me error this could be
when ran with --norc there is no mismatch ( all fine then )

no idea, .. anyone up for detailed checkings ?

i just do this fs2bash linker i have running, and yet twice in its history
i experienced files in bin/ dirs beeing cropped to one line

i suspect both have their fingers from the wrong side of the force in

peace, cheers

On Sun, Nov 28, 2021, 6:23 PM Alex fxmbsw7 Ratchev <fxmbsw7@gmail.com>
wrote:

> this is on termux's bash
> GNU bash, version 5.1.8(1)-release (aarch64-unknown-linux-android)
>
> this on updated debian also
>
> GNU bash, version 5.1.12(1)-release (aarch64-unknown-linux-gnu)
>
>
>
> On Sun, Nov 28, 2021, 6:13 PM Lawrence Velázquez <vq@larryv.me> wrote:
>
>> On Sun, Nov 28, 2021, at 3:52 AM, Alex fxmbsw7 Ratchev wrote:
>> > declare 's=1 foo["$s"]=asd' ; declare -p s foo ; echo $s ; echo "$s" ;
>> echo
>> > fin
>> >
>> > declare -- s="1 foo[\"\$s\"]=asd"
>> > bash: declare: foo: not found
>> >
>> > 1 foo["$s"]=asd
>> > fin
>> >
>> >
>> > notice the empty line
>>
>> I don't see one.
>>
>>     $ printf '%s\n' "$BASH_VERSION"
>>     5.1.8(1)-release
>>     $ unset s foo
>>     $ declare 's=1 foo["$s"]=asd' ; declare -p s foo ; echo $s ; echo
>> "$s" ; echo fin
>>     declare -- s="1 foo[\"\$s\"]=asd"
>>     bash: declare: foo: not found
>>     1 foo["$s"]=asd
>>     1 foo["$s"]=asd
>>     fin
>>
>> --
>> vq
>>
>


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