current community

your communities

more stack exchange communities

company blog
Stack Exchange Inbox Reputation and Badges
tour help
Unix & Linux Stack Exchange is a question and answer site for users of Linux, FreeBSD and other Un*x-like operating systems. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
up vote 2 down vote favorite

Suppose:

a=b; b=c; c=d

Then eval echo \$$a produces the output:

c

If we want to extract the output d using just input a, I tried the following way:

(i) eval echo \$(eval echo \$$a) produces the error:

syntax error near unexpected token '('

(ii) eval echo \$\(eval echo \$$a\) produces the output:

c

I am not able to understand why escape slashing the bracket got rid of the error.

Also, could someone please explain why I didn't get the output as d in the second instance?

share|improve this question
    
See Eval command and security issues – Wildcard yesterday
    
@Wildcard I am new to bash shell programming. The link which you provided is a bit tough to me. Could you please explain in simple words where I went wrong? – Wanderer yesterday
    
Sure. From a security standpoint, it's a really bad idea to use eval in any shell script unless you know exactly what you're doing. (And even then, there are virtually zero instances where it is actually the best solution.) As a beginner to shell scripting, please just forget that eval even exists. :) – Wildcard yesterday
    
@Wildcard ohk.. I understand but I would be really grateful if you could pin point where I went wrong in my code. – Wanderer yesterday
up vote 4 down vote accepted

First, a word of caution:

From a security standpoint, it's a really bad idea to use eval in any shell script unless you know exactly what you're doing. (And even then, there are virtually zero instances where it is actually the best solution.) As a beginner to shell scripting, please just forget that eval even exists.

For further reading, see Eval command and security issues.

To get the output d, you could use:

eval echo \$${!a}

Or:

eval eval echo \\\$\$$a

Where you went wrong was in passing the unescaped parentheses characters to echo. If they are preceded by an unquoted $, it is command substitution. But if the $ is quoted and not the parentheses, it isn't valid shell syntax.

share|improve this answer
up vote 3 down vote

Nested evals are a messy thing.

First,

eval echo \$\(eval echo \$$a\)

causes the interpreter to evaluate the following:

echo $(eval echo $b)

Interpolating the innermost nested command (i.e. $(eval echo $b)) results in:

echo c

Which gives you the result:

c

If you want the nested evals to get evaluated correctly (and I highly recommend not using eval to at all), you have to get pretty weird. See other answers for examples.

share|improve this answer
    
Thanks. Could you also please explain why back slashing the parenthesis was important here – Wanderer yesterday
up vote 1 down vote

You could get the expected output d with the following :

eval echo \$$(echo ${!a})

${!a} is a bash variable name expansion that gives c with your values. See bash manual.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.