Determine if a function exists in bash
BashFunctionTestingScriptingBash Problem Overview
Currently I'm doing some unit tests which are executed from bash. Unit tests are initialized, executed and cleaned up in a bash script. This script usualy contains an init(), execute() and cleanup() functions. But they are not mandatory. I'd like to test if they are or are not defined.
I did this previously by greping and seding the source, but it seemed wrong. Is there a more elegant way to do this?
Edit: The following sniplet works like a charm:
fn_exists()
{
LC_ALL=C type $1 | grep -q 'shell function'
}
Bash Solutions
Solution 1 - Bash
Like this: [[ $(type -t foo) == function ]] && echo "Foo exists"
The built-in type
command will tell you whether something is a function, built-in function, external command, or just not defined.
Additional examples:
$ LC_ALL=C type foo
bash: type: foo: not found
$ LC_ALL=C type ls
ls is aliased to `ls --color=auto'
$ which type
$ LC_ALL=C type type
type is a shell builtin
$ LC_ALL=C type -t rvm
function
$ if [ -n "$(LC_ALL=C type -t rvm)" ] && [ "$(LC_ALL=C type -t rvm)" = function ]; then echo rvm is a function; else echo rvm is NOT a function; fi
rvm is a function
Solution 2 - Bash
The builtin bash command declare
has an option -F
that displays all defined function names. If given name arguments, it will display which of those functions are exist, and if all do it will set status accordingly:
$ fn_exists() { declare -F "$1" > /dev/null; }
$ unset f
$ fn_exists f && echo yes || echo no
no
$ f() { return; }
$ fn_exist f && echo yes || echo no
yes
Solution 3 - Bash
If declare is 10x faster than test, this would seem the obvious answer.
Edit: Below, the -f
option is superfluous with BASH, feel free to leave it out. Personally, I have trouble remembering which option does which, so I just use both. -f shows functions, and -F shows function names.
#!/bin/sh
function_exists() {
declare -f -F $1 > /dev/null
return $?
}
function_exists function_name && echo Exists || echo No such function
The "-F" option to declare causes it to only return the name of the found function, rather than the entire contents.
There shouldn't be any measurable performance penalty for using /dev/null, and if it worries you that much:
fname=`declare -f -F $1`
[ -n "$fname" ] && echo Declare -f says $fname exists || echo Declare -f says $1 does not exist
Or combine the two, for your own pointless enjoyment. They both work.
fname=`declare -f -F $1`
errorlevel=$?
(( ! errorlevel )) && echo Errorlevel says $1 exists || echo Errorlevel says $1 does not exist
[ -n "$fname" ] && echo Declare -f says $fname exists || echo Declare -f says $1 does not exist
Solution 4 - Bash
Borrowing from other solutions and comments, I came up with this:
fn_exists() {
# appended double quote is an ugly trick to make sure we do get a string -- if $1 is not a known command, type does not output anything
[ `type -t $1`"" == 'function' ]
}
Used as ...
if ! fn_exists $FN; then
echo "Hey, $FN does not exist ! Duh."
exit 2
fi
It checks if the given argument is a function, and avoids redirections and other grepping.
Solution 5 - Bash
Dredging up an old post ... but I recently had use of this and tested both alternatives described with :
test_declare () {
a () { echo 'a' ;}
declare -f a > /dev/null
}
test_type () {
a () { echo 'a' ;}
type a | grep -q 'is a function'
}
echo 'declare'
time for i in $(seq 1 1000); do test_declare; done
echo 'type'
time for i in $(seq 1 100); do test_type; done
this generated :
real 0m0.064s
user 0m0.040s
sys 0m0.020s
type
real 0m2.769s
user 0m1.620s
sys 0m1.130s
declare is a helluvalot faster !
Solution 6 - Bash
It boils down to using 'declare' to either check the output or exit code.
Output style:
isFunction() { [[ "$(declare -Ff "$1")" ]]; }
Usage:
isFunction some_name && echo yes || echo no
However, if memory serves, redirecting to null is faster than output substitution (speaking of, the awful and out-dated `cmd` method should be banished and $(cmd) used instead.) And since declare returns true/false if found/not found, and functions return the exit code of the last command in the function so an explicit return is usually not necessary, and since checking the error code is faster than checking a string value (even a null string):
Exit status style:
isFunction() { declare -Ff "$1" >/dev/null; }
That's probably about as succinct and benign as you can get.
Solution 7 - Bash
Testing different solutions:
#!/bin/bash
test_declare () {
declare -f f > /dev/null
}
test_declare2 () {
declare -F f > /dev/null
}
test_type () {
type -t f | grep -q 'function'
}
test_type2 () {
[[ $(type -t f) = function ]]
}
funcs=(test_declare test_declare2 test_type test_type2)
test () {
for i in $(seq 1 1000); do $1; done
}
f () {
echo 'This is a test function.'
echo 'This has more than one command.'
return 0
}
post='(f is function)'
for j in 1 2 3; do
for func in ${funcs[@]}; do
echo $func $post
time test $func
echo exit code $?; echo
done
case $j in
1) unset -f f
post='(f unset)'
;;
2) f='string'
post='(f is string)'
;;
esac
done
outputs e.g.:
> test_declare (f is function) > > real 0m0,055s user 0m0,041s sys 0m0,004s exit code 0 > > test_declare2 (f is function) > > real 0m0,042s user 0m0,022s sys 0m0,017s exit code 0 > > test_type (f is function) > > real 0m2,200s user 0m1,619s sys 0m1,008s exit code 0 > > test_type2 (f is function) > > real 0m0,746s user 0m0,534s sys 0m0,237s exit code 0 > > test_declare (f unset) > > real 0m0,040s user 0m0,029s sys 0m0,010s exit code 1 > > test_declare2 (f unset) > > real 0m0,038s user 0m0,038s sys 0m0,000s exit code 1 > > test_type (f unset) > > real 0m2,438s user 0m1,678s sys 0m1,045s exit code 1 > > test_type2 (f unset) > > real 0m0,805s user 0m0,541s sys 0m0,274s exit code 1 > > test_declare (f is string) > > real 0m0,043s user 0m0,034s sys 0m0,007s exit code 1 > > test_declare2 (f is string) > > real 0m0,039s user 0m0,035s sys 0m0,003s exit code 1 > > test_type (f is string) > > real 0m2,394s user 0m1,679s sys 0m1,035s exit code 1 > > test_type2 (f is string) > > real 0m0,851s user 0m0,554s sys 0m0,294s exit code 1
So declare -F f
seems to be the best solution.
Solution 8 - Bash
From my comment on another answer (which I keep missing when I come back to this page)
$ fn_exists() { test x$(type -t $1) = xfunction; }
$ fn_exists func1 && echo yes || echo no
no
$ func1() { echo hi from func1; }
$ func1
hi from func1
$ fn_exists func1 && echo yes || echo no
yes
Solution 9 - Bash
Invocation of a function if defined.
Known function name. Let's say the name is my_function
, then use
[[ "$(type -t my_function)" == 'function' ]] && my_function;
# or
[[ "$(declare -fF my_function)" ]] && my_function;
Function's name is stored in a variable. If we declare func=my_function
, then we can use
[[ "$(type -t $func)" == 'function' ]] && $func;
# or
[[ "$(declare -fF $func)" ]] && $func;
The same results with ||
instead of &&
(Such a logic inversion could be useful during coding)
[[ "$(type -t my_function)" != 'function' ]] || my_function;
[[ ! "$(declare -fF my_function)" ]] || my_function;
func=my_function
[[ "$(type -t $func)" != 'function' ]] || $func;
[[ ! "$(declare -fF $func)" ]] || $func;
Strict mode and precondition checks
We have set -e
as a strict mode.
We use || return
in our function in a precondition.
This forces our shell process to be terminated.
# Set a strict mode for script execution. The essence here is "-e"
set -euf +x -o pipefail
function run_if_exists(){
my_function=$1
[[ "$(type -t $my_function)" == 'function' ]] || return;
$my_function
}
run_if_exists non_existing_function
echo "you will never reach this code"
The above is an equivalent of
set -e
function run_if_exists(){
return 1;
}
run_if_exists
which kills your process.
Use || { true; return; }
instead of || return;
in preconditions to fix this.
[[ "$(type -t my_function)" == 'function' ]] || { true; return; }
Solution 10 - Bash
This tells you if it exists, but not that it's a function
fn_exists()
{
type $1 >/dev/null 2>&1;
}
Solution 11 - Bash
fn_exists()
{
[[ $(type -t $1) == function ]] && return 0
}
update
isFunc ()
{
[[ $(type -t $1) == function ]]
}
$ isFunc isFunc
$ echo $?
0
$ isFunc dfgjhgljhk
$ echo $?
1
$ isFunc psgrep && echo yay
yay
$
Solution 12 - Bash
I would improve it to:
fn_exists()
{
type $1 2>/dev/null | grep -q 'is a function'
}
And use it like this:
fn_exists test_function
if [ $? -eq 0 ]; then
echo 'Function exists!'
else
echo 'Function does not exist...'
fi
Solution 13 - Bash
I particularly liked solution from Grégory Joseph
But I've modified it a little bit to overcome "double quote ugly trick":
function is_executable()
{
typeset TYPE_RESULT="`type -t $1`"
if [ "$TYPE_RESULT" == 'function' ]; then
return 0
else
return 1
fi
}
Solution 14 - Bash
It is possible to use 'type' without any external commands, but you have to call it twice, so it still ends up about twice as slow as the 'declare' version:
test_function () {
! type -f $1 >/dev/null 2>&1 && type -t $1 >/dev/null 2>&1
}
Plus this doesn't work in POSIX sh, so it's totally worthless except as trivia!