On Fri, Jun 24, 2011 at 10:38, BX <daide...@gmail.com> wrote:
> #/bin/bash
> # file1
> import()
> {
> source file2
> }
>
> import
>
> echo v1=$v1, v2=$v2, v3=$v3
>
> #/bin/bash
> # file2
> v1=v1
> declare -x v2=v2
> declare v3=v3
>
> Run above script by
> $ bash file1
>
> Expected result: v1=v1, v2=v2, v3=v3
> Real result: v1=v1, v2=, v3=
>From the documentation here:
info '(bash)Shell Functions'
we have:
When the name of a shell function is used as a
simple command name, the list of commands
associated with that function name is executed.
Shell functions are executed in the current shell
context; no new process is created to interpret
them.
...
The BODY of the function is the compound
command COMPOUND-COMMAND
and from:
info '(bash)Command Grouping'
we have:
Placing a list of commands between curly braces
causes the list to be executed in the current
shell context. No subshell is created.
and from:
info '(bash)Bourne Shell Builtins'
we know that the `source' builtin works as follows:
Read and execute commands from the FILENAME
argument in the current shell context
and from:
info '(bash)Bash Builtins'
we have
When used in a function, `declare' makes each
NAME local, as with the `local' command, unless
the `-g' option is used.
So, basically, the results are as expected: You execute the function
`import', which executes `source file2' in the current shell context,
which creates the variable `v1' in the current shell context, but
creates the variables `v2' and `v3' local to the function being
executed in the current shell context, so that by the echo statement,
only `v1' is defined.
Sincerely,
Michael Witten
