After reading the Bash man pages and with respect to this post, I am still having trouble understanding what exactly the eval command does and which would be its typical uses.
For example, if we do:
$ set -- one two three # Sets $1 $2 $3
$ echo $1
one
$ n=1
$ echo ${$n} ## First attempt to echo $1 using brackets fails
bash: ${$n}: bad substitution
$ echo $($n) ## Second attempt to echo $1 using parentheses fails
bash: 1: command not found
$ eval echo \${$n} ## Third attempt to echo $1 using 'eval' succeeds
one
What exactly is happening here and how do the dollar sign and the backslash tie into the problem?
eval takes a string as its argument, and evaluates it as if you'd typed that string on a command line. (If you pass several arguments, they are first joined with spaces between them.)
${$n} is a syntax error in bash. Inside the braces, you can only have a variable name, with some possible prefix and suffixes, but you can't have arbitrary bash syntax and in particular you can't use variable expansion. There is a way of saying “the value of the variable whose name is in this variable”, though:
echo ${!n}
one
$(…) runs the command specified inside the parentheses in a subshell (i.e. in a separate process that inherits all settings such as variable values from the current shell), and gathers its output. So echo $($n) runs $n as a shell command, and displays its output. Since $n evaluates to 1, $($n) attempts to run the command 1, which does not exist.
eval echo \${$n} runs the parameters passed to eval. After expansion, the parameters are echo and ${1}. So eval echo \${$n} runs the command echo ${1}.
Note that most of the time, you must use double quotes around variable substitutions and command substitutions (i.e. anytime there's a $): "$foo", "$(foo)". Always put double quotes around variable and command substitutions, unless you know you need to leave them off. Without the double quotes, the shell performs field splitting (i.e. it splits value of the variable or the output from the command into separate words) and then treats each word as a wildcard pattern. For example:
$ ls
file1 file2 otherfile
$ set -- 'f* *'
$ echo "$1"
f* *
$ echo $1
file1 file2 file1 file2 otherfile
$ n=1
$ eval echo \${$n}
file1 file2 file1 file2 otherfile
$eval echo \"\${$n}\"
f* *
$ echo "${!n}"
f* *
eval is not used very often. In some shells, the most common use is to obtain the value of a variable whose name is not known until runtime. In bash, this is not necessary thanks to the ${!VAR} syntax. eval is still useful when you need to construct a longer command containing operators, reserved words, etc.
Simply think of eval as "evaluating your expression one additional time before execution"
eval echo \${$n} becomes echo $1 after the first round of evaluation. Three changes to notice:
The \$ became $ (The backslash is needed, otherwise it tries to evaluate ${$n}, which means a variable named {$n}, which is not allowed)
$n was evaluated to 1
The eval disappeared
In the second round, it is basically echo $1 which can be directly executed.
So eval <some command> will first evaluate <some command> (by evaluate here I mean substitute variables, replace escaped characters with the correct ones etc.), and then run the resultant expression once again.
eval is used when you want to dynamically create variables, or to read outputs from programs specifically designed to be read like this. See Eval command and security issues for examples. The link also contains some typical ways in which eval is used, and the risks associated with it.
In my experience, a "typical" use of eval is for running commands that generate shell commands to set environment variables.
Perhaps you have a system that uses a collection of environment variables, and you have a script or program that determines which ones should be set and their values. Whenever you run a script or program, it runs in a forked process, so anything it does directly to environment variables is lost when it exits. But that script or program can send the export commands to standard output.
Without eval, you would need to redirect standard output to a temporary file, source the temporary file, and then delete it. With eval, you can just:
eval "$(script-or-program)"
Note the quotes are important. Take this (contrived) example:
# activate.sh
echo 'I got activated!'
# test.py
print("export foo=bar/baz/womp")
print(". activate.sh")
$ eval $(python test.py)
bash: export: `.': not a valid identifier
bash: export: `activate.sh': not a valid identifier
$ eval "$(python test.py)"
I got activated!
The eval statement tells the shell to take eval’s arguments as commands and run them through the command-line. It is useful in a situation like below:
In your script if you are defining a command into a variable and later on you want to use that command then you should use eval:
a="ls | more"
$a
Output:
bash: command not found: ls | more
The above command didn't work as ls tried to list file with name pipe (|) and more. But these files are not there:
eval $a
Output:
file.txt
mailids
remote_cmd.sh
sample.txt
tmp
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.
I originally intentionally never learned how to use eval, because most people will recommend to stay away from it like the plague. However I recently discovered a use case that made me facepalm for not recognizing it sooner.
If you have cron jobs that you want to run interactively to test, you might view the contents of the file with cat, and copy and paste the cron job to run it. Unfortunately, this involves touching the mouse, which is a sin in my book.
Lets say you have a cron job at /etc/cron.d/repeatme with the contents:
*/10 * * * * root program arg1 arg2
You cant execute this as a script with all the junk in front of it, but we can use cut to get rid of all the junk, wrap it in a subshell, and execute the string with eval
eval $( cut -d ' ' -f 6- /etc/cron.d/repeatme)
The cut command only prints out the 6th field of the file, delimited by spaces. Eval then executes that command.
I used a cron job here as an example, but the concept is to format text from stdout, and then evaluate that text.
The use of eval in this case is not insecure, because we know exactly what we will be evaluating before hand.
I've recently had to use eval to force multiple brace expansions to be evaluated in the order I needed. Bash does multiple brace expansions from left to right, so
xargs -I_ cat _/{11..15}/{8..5}.jpg
expands to
xargs -I_ cat _/11/8.jpg _/11/7.jpg _/11/6.jpg _/11/5.jpg _/12/8.jpg _/12/7.jpg _/12/6.jpg _/12/5.jpg _/13/8.jpg _/13/7.jpg _/13/6.jpg _/13/5.jpg _/14/8.jpg _/14/7.jpg _/14/6.jpg _/14/5.jpg _/15/8.jpg _/15/7.jpg _/15/6.jpg _/15/5.jpg
but I needed the second brace expansion done first, yielding
xargs -I_ cat _/11/8.jpg _/12/8.jpg _/13/8.jpg _/14/8.jpg _/15/8.jpg _/11/7.jpg _/12/7.jpg _/13/7.jpg _/14/7.jpg _/15/7.jpg _/11/6.jpg _/12/6.jpg _/13/6.jpg _/14/6.jpg _/15/6.jpg _/11/5.jpg _/12/5.jpg _/13/5.jpg _/14/5.jpg _/15/5.jpg
The best I could come up with to do that was
xargs -I_ cat $(eval echo _/'{11..15}'/{8..5}.jpg)
This works because the single quotes protect the first set of braces from expansion during the parsing of the eval command line, leaving them to be expanded by the subshell invoked by eval.
There may be some cunning scheme involving nested brace expansions that allows this to happen in one step, but if there is I'm too old and stupid to see it.
You asked about typical uses.
One common complaint about shell scripting is that you (allegedly) can't pass by reference to get values back out of functions.
But actually, via "eval", you can pass by reference. The callee can pass back a list of variable assignments to be evaluated by the caller. It is pass by reference because the caller can allowed to specify the name(s) of the result variable(s) - see example below. Error results can be passed back standard names like errno and errstr.
Here is an example of passing by reference in bash:
#!/bin/bash
isint()
{
re='^[-]?[0-9]+$'
[[ $1 =~ $re ]]
}
#args 1: name of result variable, 2: first addend, 3: second addend
iadd()
{
if isint ${2} && isint ${3} ; then
echo "$1=$((${2}+${3}));errno=0"
return 0
else
echo "errstr=\"Error: non-integer argument to iadd $*\" ; errno=329"
return 1
fi
}
var=1
echo "[1] var=$var"
eval $(iadd var A B)
if [[ $errno -ne 0 ]]; then
echo "errstr=$errstr"
echo "errno=$errno"
fi
echo "[2] var=$var (unchanged after error)"
eval $(iadd var $var 1)
if [[ $errno -ne 0 ]]; then
echo "errstr=$errstr"
echo "errno=$errno"
fi
echo "[3] var=$var (successfully changed)"
The output looks like this:
[1] var=1
errstr=Error: non-integer argument to iadd var A B
errno=329
[2] var=1 (unchanged after error)
[3] var=2 (successfully changed)
There is almost unlimited band width in that text output! And there are more possibilities if the multiple output lines are used: e.g., the first line could be used for variable assignments, the second for continuous 'stream of thought', but that's beyond the scope of this post.
In the question:
who | grep $(tty | sed s:/dev/::)
outputs errors claiming that files a and tty do not exist. I understood this to mean that tty is not being interpreted before execution of grep, but instead that bash passed tty as a parameter to grep, which interpreted it as a file name.
There is also a situation of nested redirection, which should be handled by matched parentheses which should specify a child process, but bash is primitively a word separator, creating parameters to be sent to a program, therefore parentheses are not matched first, but interpreted as seen.
I got specific with grep, and specified the file as a parameter instead of using a pipe. I also simplified the base command, passing output from a command as a file, so that i/o piping would not be nested:
grep $(tty | sed s:/dev/::) <(who)
works well.
who | grep $(echo pts/3)
is not really desired, but eliminates the nested pipe and also works well.
In conclusion, bash does not seem to like nested pipping. It is important to understand that bash is not a new-wave program written in a recursive manner. Instead, bash is an old 1,2,3 program, which has been appended with features. For purposes of assuring backward compatibility, the initial manner of interpretation has never been modified. If bash was rewritten to first match parentheses, how many bugs would be introduced into how many bash programs? Many programmers love to be cryptic.
As clearlight has said, "(p)erhaps the most straightforward example would be a bash program that opens another bash script as a text file, reads each line of text, and uses eval to execute them in order". I'm no expert, but the textbook I'm currently reading (Shell-Programmierung by Jürgen Wolf) points to one particular use of this that I think would be a valuable addition to the set of potential use cases collected here.
For debugging purposes, you may want to go through your script line by line (pressing Enter for each step). You could use eval to execute every line by trapping the DEBUG signal (which I think is sent after every line):
trap 'printf "$LINENO :-> " ; read line ; eval $line' DEBUG
I like the "evaluating your expression one additional time before execution" answer, and would like to clarify with another example.
var="\"par1 par2\""
echo $var # prints nicely "par1 par2"
function cntpars() {
echo " > Count: $#"
echo " > Pars : $*"
echo " > par1 : $1"
echo " > par2 : $2"
if [[ $# = 1 && $1 = "par1 par2" ]]; then
echo " > PASS"
else
echo " > FAIL"
return 1
fi
}
# Option 1: Will Pass
echo "eval \"cntpars \$var\""
eval "cntpars $var"
# Option 2: Will Fail, with curious results
echo "cntpars \$var"
cntpars $var
The curious results in option 2 are that we would have passed two parameters as follows:
First parameter: "par1
Second parameter: par2"
How is that for counter intuitive? The additional eval will fix that.
It was adapted from another answer on How can I reference a file for variables using Bash?
There is a file called test.txt that contains:
ljlkfjdslkfldjfdsajflkjf word:test1
dflkdjflkdfdjls word:test2
dlkdj word:test3
word:test4
word:NewYork
dljfldflkdflkdjf word:test7
djfkd word:young
dkjflke word:lisa
amazonwle word:NewYork
dlksldjf word:test10
Now all we want is to get the strings after colon and if the result is same, print the output, in this case it is "NewYork"
Here is the script which lists the elements but when tried to push into array and compare it is failing, Please let me know my mistakes.
#!/usr/bin/sh
input="test.txt"
cat $input | while read line; do output= $(echo $line | cut -d":" -f2); done
for (( i = 0 ; i < "${#output[#]}" ; i++ ))
{
echo ${output[i]}
}
Error obtained:
./compare.sh
./compare.sh: line 11: test1: command not found
./compare.sh: line 11: test2: command not found
./compare.sh: line 11: test3: command not found
./compare.sh: line 11: test4: command not found
./compare.sh: line 11: NewYork: command not found
./compare.sh: line 11: raghav: command not found
./compare.sh: line 11: young: command not found
./compare.sh: line 11: lisa: command not found
./compare.sh: line 11: NewYork: command not found
./compare.sh: line 11: test10: command not found
Please let me know my mistakes.
output= $(..) first excutes the command $(..) inside and grabs it's output. It then sets the variable output to an empty string as-if the same as output="" and exports the variable output and executes the output of $(..) as a command. Remove the space after =.
You are setting output on the right side of a pipe inside a subshell. The changes will not be visible outside - output is unset once the pipe terminates. Use redirection while .... done < file.
And output is not an array, but a normal variable. There is no ${output[i]} (well, except ${output[0]}) as it's not an array (and output is unset, as explained above). Append an element to an array output+=("$(...)").
#!/usr/bin/sh - is invalid, sh may not be bash and support bash arrays. To use bash extension specifically use bash and use a shebang with bash - #!/bin/bash,
Now stylistic:
The format for .... { ... } is supported in bash since forever, however it's a rather undocumented syntax rather to be deprecated to be readable by people used to programming in C. Prefer the standard do ... done.
The expansions $input and ${output[i]} are unquoted.
The read will ignore leading and trailing whitespaces and also interpret/ignore \ backslash sequences (ie. see -r option to read).
echo $line will make $line undergo word splitting - multiple whitespaces will be replaced by a single space.
And using cut after read can be simpler written as just read with IFS=: and splitting on read. Also, if you're using cut, you could just cut -d: -f2 file on the whole file instead of cutting one line at a time.
Re-read basic bash introduction - notice that bash is space aware, spaces around = count. Read bashfaq how to read a file field by field, read about subshells and environments, read bashfaq I set variables in a loop that's in a pipeline. Why do they disappear after the loop terminates? Or, why can't I pipe data to read? and an introduction to bash arrays.
I am trying to write a Bash script that appends a string to a Bash array, where the string contains the path to a Python script together with the arguments passed into the Bash script, enclosed in double quotes.
If I call the script using ./script.sh -o "a b", I would like a CMD_COUNT of 1, but I am getting 2 instead.
script.sh:
#!/bin/bash
declare -a COMMANDS=()
COMMANDS+=("/path/to/myscript.py \"${#}\"")
CMD_COUNT=${#COMMANDS[*]}
echo $CMD_COUNT
How can I ensure that the appended string is /path/to/myscript.py "-o" "a b"?
EDIT: The full script is actually like this:
script.sh:
#!/bin/bash
declare -a COMMANDS=()
COMMANDS+=("/path/to/myscript2.py")
COMMANDS+=("/path/to/myscript.py \"${#}\"")
CMD_COUNT=${#COMMANDS[*]}
echo $CMD_COUNT
for i in ${!COMMANDS[*]}
do
echo "${0} - command: ${COMMANDS[${i}]}"
${COMMANDS[${i}]}
done
It's a bad idea, but if it's what you really want, printf %q can be used to generate a string that, when parsed by the shell, will result in a given list of arguments. (The exact escaping might not be identical to what you'd write by hand, but the effect of evaluating it -- using eval -- will be).
#!/bin/bash
declare -a COMMANDS=( )
printf -v command '%q ' "/path/to/myscript" "$#"
COMMANDS+=( "$command" )
CMD_COUNT=${#COMMANDS[#]}
echo "$CMD_COUNT"
...but, as I said, this is all a bad idea.
Best-practice ways to encapsulate code as data in bash involve using functions, or arrays with one element per argument.
eval results in code that's prone to security bugs.
I'm writing a small script in which I want to set the value of a variable equal to the output of a command. However, the command in question is a call to another script with command-line arguments. I'm using backticks as you normally should in this scenario, but the problem is that the the computer gives an error, in which it tries to interpret the command-line arguments as commands.
#!/bin/bash
filename="$1"
while read p; do
echo "This is the gene we are looking at: ""$p"
lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
echo "$lookIn"
#grep "$p" "$lookIn""/""prokka_""$lookIn""/*.tsv" | awk '{print $1}'
done < $filename
I'm trying to set variable lookIn equal to the output of ./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri, where ./findGeneIn is a script, and the words burgdorferi,...,parkeri are command line arguments for ./findGeneIn.
The issue, is that I get an error saying "burgdorferi: command not found". So it's trying to interpret those arguments as commands. How do I get it to not do that?
lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
^
Delete the extra space. Assignments must not have spaces around the equal sign.
With the space there, Bash parses the line as var=value command, which runs a command with $var temporarily set to "value". Or in this case, it interprets result of the backticks as a command name and lookIn= as an empty variable assignment.
I intend to read the lines of a short .txt file and assign each line to variables containing the line number in the variable name.
File example.txt looks like this:
Line A
Line B
When I run the following code:
i=1
while read line; do
eval line$i="$line"
echo $line
((i=i+1))
done < example.txt
What I would expect during execution is:
Line A
Line B
and afterwards being able to call
$ echo $line1
Line A
$ echo $line2
Line B
However, the code above results in the error:
-bash: A: command not found
Any ideas for a fix?
Quote-removal happens twice with eval. Your double-quotes are getting removed before eval even runs. I'm not even going to directly answer that part, because there are better ways to do this:
readarray line < example.txt # bash 4
echo "${line[0]}"
Or, to do exactly what you were doing, with a different variable for each line:
i=1
while read line$((i++)); do
:
done < example.txt
Also check out printf -v varname "%s" value for a better / safer way to assign by reference.
Check out the bash-completion code if you want to see some complicated call-by-reference bash shenanigans.
Addressing your comment: if you want to process lines as they come in, but still save previous lines, I'd go with this construct:
lines=()
while read l;do
lines+=( "$l" )
echo "my line is $l"
done < "$infile"
This way you don't have to jump through any syntactic hoops to access the current line (vs. having to declare a reference-variable to line$i, or something.)
Bash arrays are really handy, because you can access a single element by value, or you can do "${#lines[#]}" to get the line count. Beware that unset lines[4] leaves a gap, rather than renumbering lines[5:infinity]. See the "arrays" section in the bash man page. To find the part of the manual that documents $# expansion, and other stuff, search in the manual for ##. The Parameter Expansion section is the first hit for that in the bash 4.3 man page.
eval line$i="$line" tells bash to evaluate the string "line1=Line A", which attempts to invoke a command named A with the environment variable "line1" set to the value of Line. You probably want to do eval "line$i='$line'"