How to combine timeout and eval commands in bash

Question

For executing command that is stored in variable the eval command is used:

└──> a="echo -e 'a\nb' | wc -l"
└──> eval $a
2

But how can it be combined with timeout command? I've tried following which gives me wrong output:

└──> timeout 10 $a
'a
b' | wc -l

And the following which gives me errors:

└──> timeout 10 "$a"
timeout: failed to run command `echo -e \'a\\nb\' | wc -l': No such file or directory

└──> timeout 10 $(eval $a)
timeout: failed to run command `2': No such file or directory

└──> timeout 10 $(eval "$a")
timeout: failed to run command `2': No such file or directory

The question can also stand: How can I be sure that following command is executed properly?

timeout 10 "$PROGRAM" "$OPT1" "$OPT2" ...

Don't store commands in variables. Use functions instead. You can re-declare function, just as you can re-assign variables. — anishsane
– anishsane, Commented May 12, 2015 at 10:59
& To answer your question, you can run eval "timeout 10 $a". But always remember that eval is evil & use of eval is always discouraged, unless there is absolutely no other way. — anishsane
– anishsane, Commented May 12, 2015 at 11:01
Do I even need eval when my PROGRAM="echo" OPT1="Hello"? I thing the eval is useful when there are pipes (as mentioned in original question) or some other awkward characters. — Wakan Tanka
– Wakan Tanka, Commented May 12, 2015 at 11:33
@anishane why is eval evil? your argument is similar to saying: ls is evil because for i in $(ls) is bad. Only bad usages of eval are bad. Now, I agree that here, OP is using an antipattern (code is not data in shell); but in the absolute, eval is no more evil than other commands. Also, there are very good usages of eval, and I don't see why its use should be discouraged. Bad usages should be discouraged, but that's a tautology. — gniourf_gniourf
– gniourf_gniourf, Commented Feb 11, 2016 at 12:50

agc · Accepted Answer · 2020-03-31 12:03:44Z

4

Simple:

a="echo -e 'a\nb' | wc -l"
eval timeout 10 $a

Output:

edited Mar 31, 2020 at 12:03

answered Feb 11, 2019 at 3:40

agc

8,5342 gold badges33 silver badges53 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Denis Tsoi Over a year ago

add an explanation to improve answer quality

agc Over a year ago

@DenisTsoi, Explanations can be helpful, but here I dunno... what is it about this short answer that needs explaining? It's just two commands, a parameter, and the OP's $a string.

GolDDranks Over a year ago

Note that if the command includes bash internals, such as just declaring variables, this eval timeout will fail, because timeout can't process those commands, it just calls an external command in a subshell.

RTLinuxSW · Accepted Answer · 2015-05-12 19:38:25Z

1

This will work

if timeout "$PROGRAM" "$OPT1" "$OPT2" ... ; then
    echo Program ran successfully
else
    echo Program terminated due to timeout
fi

answered May 12, 2015 at 19:38

RTLinuxSW

8525 silver badges9 bronze badges

Comments

user1528305 · Accepted Answer · 2016-02-11 12:16:11Z

0

If it's about keeping commands in variables, this will work, although don't know if it's a 'proper bash way' to do it:

command.sh:

#!/bin/bash
echo -e 'a\nb' | wc -l

run.sh:

#!/bin/bash
a="command.sh"
timeout 10 ./$a

answered Feb 11, 2016 at 12:16

user1528305

Comments

Community · Accepted Answer · 2017-05-23 10:33:48Z

0

echo "$(timeout 10 echo -e 'a\nb' | wc -l)"
OR
echo "$(timeout 2 echo "$(eval $a)")"

explanation 1 here : HERE

explanation 2 here : HERE

edited May 23, 2017 at 10:33

CommunityBot

11 silver badge

answered May 20, 2016 at 8:40

diveinsky

1652 silver badges8 bronze badges

Collectives™ on Stack Overflow

How to combine timeout and eval commands in bash

4 Answers 4

3 Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

3 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related