Update: Some people say one should -never- use eval. I disagree. I think the risk arises when corrupt input can be passed to eval. However there are many common situations where that is not a risk, and therefore it is worth knowing how to use eval in any case. This stackoverflow answer explains the risks of eval and alternatives to eval. Ultimately it is up to the user to determine if/when eval is safe and efficient to use.
The bash eval statement allows you to execute lines of code calculated or acquired, by your bash script.
Perhaps the most straightforward example would be a bash program that opens another bash script as a text file, reads each line of text, and uses eval to execute them in order. That's essentially the same behavior as the bash source statement, which is what one would use, unless it was necessary to perform some kind of transformation (e.g. filtering or substitution) on the content of the imported script.
I rarely have needed eval, but I have found it useful to read or write variables whose names were contained in strings assigned to other variables. For example, to perform actions on sets of variables, while keeping the code footprint small and avoiding redundancy.
eval is conceptually simple. However, the strict syntax of the bash language, and the bash interpreter's parsing order can be nuanced and make eval appear cryptic and difficult to use or understand. Here are the essentials:
The argument passed to eval is a string expression that is calculated at runtime. eval will execute the final parsed result of its argument as an actual line of code in your script.
Syntax and parsing order are stringent. If the result isn't an executable line of bash code, in scope of your script, the program will crash on the eval statement as it tries to execute garbage.
When testing you can replace the eval statement with echo and look at what is displayed. If it is legitimate code in the current context, running it through eval will work.
The following examples may help clarify how eval works...
Example 1:
eval statement in front of 'normal' code is a NOP
$ eval a=b
$ eval echo $a
b
In the above example, the first eval statements has no purpose and can be eliminated. eval is pointless in the first line because there is no dynamic aspect to the code, i.e. it already parsed into the final lines of bash code, thus it would be identical as a normal statement of code in the bash script. The 2nd eval is pointless too, because, although there is a parsing step converting $a to its literal string equivalent, there is no indirection (e.g. no referencing via string value of an actual bash noun or bash-held script variable), so it would behave identically as a line of code without the eval prefix.
Example 2:
Perform var assignment using var names passed as string values.
$ key="mykey"
$ val="myval"
$ eval $key=$val
$ echo $mykey
myval
If you were to echo $key=$val, the output would be:
mykey=myval
That, being the final result of string parsing, is what will be executed by eval, hence the result of the echo statement at the end...
Example 3:
Adding more indirection to Example 2
$ keyA="keyB"
$ valA="valB"
$ keyB="that"
$ valB="amazing"
$ eval eval \$$keyA=\$$valA
$ echo $that
amazing
The above is a bit more complicated than the previous example, relying more heavily on the parsing-order and peculiarities of bash. The eval line would roughly get parsed internally in the following order (note the following statements are pseudocode, not real code, just to attempt to show how the statement would get broken down into steps internally to arrive at the final result).
eval eval \$$keyA=\$$valA # substitution of $keyA and $valA by interpreter
eval eval \$keyB=\$valB # convert '$' + name-strings to real vars by eval
eval $keyB=$valB # substitution of $keyB and $valB by interpreter
eval that=amazing # execute string literal 'that=amazing' by eval
If the assumed parsing order doesn't explain what eval is doing enough, the third example may describe the parsing in more detail to help clarify what is going on.
Example 4:
Discover whether vars, whose names are contained in strings, themselves contain string values.
a="User-provided"
b="Another user-provided optional value"
c=""
myvarname_a="a"
myvarname_b="b"
myvarname_c="c"
for varname in "myvarname_a" "myvarname_b" "myvarname_c"; do
eval varval=\$$varname
if [ -z "$varval" ]; then
read -p "$varname? " $varname
fi
done
In the first iteration:
varname="myvarname_a"
Bash parses the argument to eval, and eval sees literally this at runtime:
eval varval=\$$myvarname_a
The following pseudocode attempts to illustrate how bash interprets the above line of real code, to arrive at the final value executed by eval. (the following lines descriptive, not exact bash code):
1. eval varval="\$" + "$varname" # This substitution resolved in eval statement
2. .................. "$myvarname_a" # $myvarname_a previously resolved by for-loop
3. .................. "a" # ... to this value
4. eval "varval=$a" # This requires one more parsing step
5. eval varval="User-provided" # Final result of parsing (eval executes this)
Once all the parsing is done, the result is what is executed, and its effect is obvious, demonstrating there is nothing particularly mysterious about eval itself, and the complexity is in the parsing of its argument.
varval="User-provided"
The remaining code in the example above simply tests to see if the value assigned to $varval is null, and, if so, prompts the user to provide a value.
$($n)runs$nin a subshell. It tries to run the command1which does not exist. – Deserveecho $1eventually, not1. I don't think it can be done using subshells. – Morellieval. – Dipody