Re: Inconsistent termination of function when 'local' tries to override read-only variable

[Chet Ramey]

On 2/19/20 2:36 PM, Arfrever Frehtes Taifersar Arahesis wrote:
> When scalar variable is read-only, then calling 'local' for this
> variable (regardless of presence of value in assignment) is non-fatal
> and subsequent commands in function are executed.

Correct. This is an implementation choice, and a controversial one, and
one that's been in place for about as long as bash has had local variables.
Bash creates the local variable with the inherited value, but doesn't allow
it to be changed. The rationale is that variables are readonly for a
reason, and you shouldn't be able to override readonly variables just by
calling a function. The local variable is created, so it's not a variable
assignment error.

There are advocates on both sides of this issue.

Here's the last time the issue came up:

https://lists.gnu.org/archive/html/bug-bash/2019-03/msg00150.html

It's come up multiple times over the years.

-- 
``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/