Is it possible to use indirection for setting variables?

Question

E.g. next script will print Hello due to ${! rule.

A=B
B=Hello
echo ${!A}

But how can I set value to variable with indirect name? Straightforward use of ! sign doesn't work:

A=B
!A=Hello # here is tricky line
echo $B

I know there is a trick with using temp file, but I'm interested in using indirection, not something like

A=B
echo "$A=Hello" > 1.tmp
. 1.tmp
echo $B

Do you want to create synonyms so that both variables point to the same data? In other words when one changes, changes the other? I'm not sure I get your intention here. — Christian, Mar 18 '13 at 16:37
no, I just wanted to set one variable through value of another once — Sergey Grinev, Mar 18 '13 at 16:44

score 16 · Answer 1 · answered Mar 18 '13 at 16:31

16

You can do this with the typeset builtin:

$ A=B
$ typeset $A=42
$ echo $B
42

Or using declare:

$ declare $A=1
$ echo $B
1

answered Mar 18 '13 at 16:31

Mat

52,586

1

In bash, typeset is obsolete and its replacement is declare. They are more or less identical. – jw013 Mar 18 '13 at 16:34
1

typeset is a bit more portable though (works in ksh) – Mat Mar 18 '13 at 16:36
thanks! Can't decide which answer is better. Flipped a coin :) – Sergey Grinev Mar 18 '13 at 20:21

score 9 · Answer 2 · edited Apr 13 '17 at 12:36

Following on indirection when getting the value of a variable…

The portable way is to use eval. You do have to pay attention to the quoting so that special characters in the value are not evaluated when they shouldn't be. The easiest way is to store the new value in an intermediate variable and assign the value of that variable.

A=B
tmp='stuff with special characters or whatever…'
eval $A=\$tmp

The argument to eval is B=$tmp, a straightforward assignment. You don't need double quotes around $tmp even if it contains whitespace or globbing characters, because those are not expanded in an assignment.

If you want to make an environment variable, you can use export. In bash or ksh or zsh, you can use typeset to make a local variable (declare is synonymous in bash). Again, you should use an intermediate variable so that special characters are not mangled. Note that except in zsh, you need to put double quotes around the variable expansion, because this is not an assignment but a builtin that takes an argument that happens to look like an assignment.

export $A="$tmp"

This (using eval) is the correct answer. It's portable, adheres to the POSIX standard, and is the one used all over the place in more advanced shell scripts. One thing, though: using export on the same line as the assignment is not portable. — MadScientist, Mar 19 '13 at 12:32
@MadScientist export var=value is POSIX. It isn't portable to Bourne shells, but these are a vanishing breed (just make sure to use /usr/xpg4/bin/sh or /usr/xpg6/bin/sh or ksh and not /bin/sh on older Solaris machines). — Gilles 'SO- stop being evil', Mar 19 '13 at 12:58

Scrutinizer · Accepted Answer · 2013-03-18T20:51:14.350

5

Another way you could try:

$ A=B
$ read $A <<< 81
$ echo "$B"
81

But there is a security risk (!) as with all these methods (also declare / typeset and of course eval) .. In this case one must control the lefthand side (the value of $A), in other words, at least in bash, variable $A should not contain user controlled input, for example an input file, etcetera...

If your shell does not support a here-string ( <<< ) you could also use a here-document instead:

A=B
read $A << EOF
82
EOF
echo "$B"

edited Mar 18 '13 at 20:51

answered Mar 18 '13 at 16:43

Scrutinizer

1,142
5
7

wow, but why does it work? – Sergey Grinev Mar 18 '13 at 16:47
2

The shell expands variable $A before the read command.. – Scrutinizer Mar 18 '13 at 17:08
The only issue: it doesn't work on Solaris – Sergey Grinev Mar 18 '13 at 18:05
What version of bash on Solaris are you referring to? On what version of Solaris? – Scrutinizer Mar 18 '13 at 18:32
Solaris 10, bash from the box – Sergey Grinev Mar 18 '13 at 19:41
I just tried this in bash 3.00 on Solaris 10 and it worked. Are you certain you using bash? If so what version is it? – Scrutinizer Mar 18 '13 at 19:48
GNU bash, version 2.05.0(1)-release (sparc-sun-solaris2.4). Maybe I just have old bash, or it's Sparc Solaris issue. Tried on i586 solaris and it worked (with bash 3 although). Sorry for the confusion. – Sergey Grinev Mar 18 '13 at 19:55
1

You could try a here-document instead of a here-string, see the edit in my post.. – Scrutinizer Mar 18 '13 at 20:13
2

This is not a good method. It's complex, and it breaks if there are special characters in the value (newlines, backslashes and leading or trailing whitespace on a line are problematic). – Gilles 'SO- stop being evil' Mar 19 '13 at 00:18
This isn't a good method, but your note about controlling variable names within a program is right on. – ormaaj Mar 20 '13 at 05:37
Depending on your bash version, it may also support A=B; read -r $A < <( echo "81" ) ; echo $B All the same security points above still apply. – Jesse Chisholm May 21 '20 at 01:32

score 2 · Answer 4 · answered Jun 12 '21 at 19:23

In bash version 5 I use declare -n varref=varname. Then variable could be set as varref=123 and read as echo $varref (does the same as echo $varname) without any additional commands and syntax.

from help declare:

-n make NAME a reference to the variable named by its value

score 1 · Answer 5 · answered Nov 23 '22 at 16:20

1

Another way to do this is with printf, which assigns its output to a variable whose name may be passed with the -v parameter:

A=B
printf -v "$A" '%s' "Hello"
echo ${!A}

answered Nov 23 '22 at 16:20

Digital Trauma

8,554

Is it possible to use indirection for setting variables?

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