Panorama Média Global

Question

I'm working with this:

GNU bash, version 4.1.2(1)-release (x86_64-redhat-linux-gnu)

I have a script like below:

#!/bin/bash

e=2

function test1() {
  e=4
  echo "hello"
}

test1 
echo "$e"

Which returns:

hello
4

But if I assign the result of the function to a variable, the global variable e is not modified:

#!/bin/bash

e=2

function test1() {
  e=4
  echo "hello"
}

ret=$(test1)

echo "$ret"
echo "$e"

Returns:

hello
2

I've heard of the use of eval in this case, so I did this in test1:

eval 'e=4'

But the same result.

Could you explain me why it is not modified? How could I save the echo of the test1 function in ret and modify the global variable too?

Accepted Answer

When you use a command substitution (i.e., the $(...) construct), you are creating a subshell. Subshells inherit variables from their parent shells, but this only works one way: A subshell cannot modify the environment of its parent shell.

Your variable e is set within a subshell, but not the parent shell. There are two ways to pass values from a subshell to its parent. First, you can output something to stdout, then capture it with a command substitution:

myfunc() {
    echo "Hello"
}

var="$(myfunc)"

echo "$var"

The above outputs:

Hello

For a numerical value in the range of 0 through 255, you can use return to pass the number as the exit status:

mysecondfunc() {
    echo "Hello"
    return 4
}

var="$(mysecondfunc)"
num_var=$?

echo "$var - num is $num_var"

This outputs:

Hello - num is 4

Answer 2

This needs bash 4.1 if you use {fd} or local -n.

The rest should work in bash 3.x I hope. I am not completely sure due to printf %q - this might be a bash 4 feature.

Summary

Your example can be modified as follows to archive the desired effect:

# Add following 4 lines:
_passback() { while [ 1 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; return $1; }
passback() { _passback "$@" "$?"; }
_capture() { { out="$("${@:2}" 3<&-; "$2_" >&3)"; ret=$?; printf "%q=%q;" "$1" "$out"; } 3>&1; echo "(exit $ret)"; }
capture() { eval "$(_capture "$@")"; }

e=2

# Add following line, called "Annotation"
function test1_() { passback e; }
function test1() {
  e=4
  echo "hello"
}

# Change following line to:
capture ret test1 

echo "$ret"
echo "$e"

prints as desired:

hello
4

Note that this solution:

Works for e=1000, too.
Preserves $? if you need $?

The only bad sideffects are:

It needs a modern bash.
It forks quite more often.
It needs the annotation (named after your function, with an added _)
It sacrifices file descriptor 3.
- You can change it to another FD if you need that.
  - In _capture just replace all occurances of 3 with another (higher) number.

The following (which is quite long, sorry for that) hopefully explains, how to adpot this recipe to other scripts, too.

The problem

d() { let x++; date +%Y%m%d-%H%M%S; }

x=0
d1=$(d)
d2=$(d)
d3=$(d)
d4=$(d)
echo $x $d1 $d2 $d3 $d4

outputs

0 20171129-123521 20171129-123521 20171129-123521 20171129-123521

while the wanted output is

4 20171129-123521 20171129-123521 20171129-123521 20171129-123521

The cause of the problem

Shell variables (or generally speaking, the environment) is passed from parental processes to child processes, but not vice versa.

If you do output capturing, this usually is run in a subshell, so passing back variables is difficult.

Some even tell you, that it is impossible to fix. This is wrong, but it is a long known difficult to solve problem.

There are several ways on how to solve it best, this depends on your needs.

Here is a step by step guide on how to do it.

Passing back variables into the parental shell

There is a way to pass back variables to a parental shell. However this is a dangerous path, because this uses eval. If done improperly, you risk many evil things. But if done properly, this is perfectly safe, provided that there is no bug in bash.

_passback() { while [ 0 -lt $# ]; do printf '%q=%q;' "$1" "${!1}"; shift; done; }

d() { let x++; d=$(date +%Y%m%d-%H%M%S); _passback x d; }

x=0
eval `d`
d1=$d
eval `d`
d2=$d
eval `d`
d3=$d
eval `d`
d4=$d
echo $x $d1 $d2 $d3 $d4

prints

4 20171129-124945 20171129-124945 20171129-124945 20171129-124945

Note that this works for dangerous things, too:

danger() { danger="$*"; passback danger; }
eval `danger '; /bin/echo *'`
echo "$danger"

prints

; /bin/echo *

This is due to printf '%q', which quotes everything such, that you can re-use it in a shell context safely.

But this is a pain in the a..

This does not only look ugly, it also is much to type, so it is error prone. Just one single mistake and you are doomed, right?

Well, we are at shell level, so you can improve it. Just think about an interface you want to see, and then you can implement it.

Augment, how the shell processes things

Let's go a step back and think about some API which allows us to easily express, what we want to do.

Well, what do we want do do with the d() function?

We want to capture the output into a variable. OK, then let's implement an API for exactly this:

# This needs a modern bash 4.3 (see "help declare" if "-n" is present,
# we get rid of it below anyway).
: capture VARIABLE command args..
capture()
{
local -n output="$1"
shift
output="$("$@")"
}

Now, instead of writing

d1=$(d)

we can write

capture d1 d

Well, this looks like we haven't changed much, as, again, the variables are not passed back from d into the parent shell, and we need to type a bit more.

However now we can throw the full power of the shell at it, as it is nicely wrapped in a function.