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Full Discussion: [bash] why my variable is not updated?
Top Forums Shell Programming and Scripting [bash] why my variable is not updated? Post 303032985 by Don Cragun on Wednesday 27th of March 2019 11:55:26 PM
Old 03-28-2019
Quote:
Originally Posted by bakunin
... ... ...

The next problem is:

Code:
tot_files=$(<...>) &

This is not working at several levels.

First: what do you want to assign to the variable tot_files? Obviously the stdout of whatever is inside the subshell ( "$(...) ). It is like this:

Code:
var=$( echo "blabla" )

The shell opens a subshell and executes echo "blabla" in this subshell. So you get some output at <stdout>, in this case a string. This output is then assigned to the variable.

The next thing is you put that assignment into background:
Code:
tot_files=<something> &

But an assignment is not a separate process, it is done within the shell internally. x=5 does not generate a process you could put into background (which is in fact just another way of saying it runs asynchronously to the main process). I am not sure about that (perhaps Don Cragun knows more about these fringe situations) but i suppose that the "&" is simply ignored in this case.

... ... ...

bakunin
Actually, this is pretty straight forward. But, an "&" is never simply ignored in the shell command language. The "&" causes the previous list (where "list" in this case is a simple command that is an assignment statement) to be run asynchronously in a subshell. So, we not only have a subshell due to the parentheses, we have a sub-subshell to run the assignment statement in the background. It is not clear to me that the standards mandate whether the command substitution that is part of this command is run in the background or whether the shell can run the command substitution in the foreground and then run the assignment of the results of that command substitution to the specified variable in the background.

In either case result of the variable assignment is visible only in the sub-subshell; it is not visible in the environment of the shell that started the subshell and it is not visible in the subshell that started the asynchronous list.

I hope this helps.
- Don
These 2 Users Gave Thanks to Don Cragun For This Post:
bakunin soichiro
 

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