Storing a function inside a bash array element

Question

How can I define a function as an element of an array?

I know I can do this:

function array[5](){ echo "you are inside the function"; }

And it is a valid function, in the sense that, the instruction array[5], correctly execute the function.

Unlucky, the function is not listed in the keys of the array, so, echo ${!array[*]} does not return 5 as a key.

For instance, the following code:

array[0]="first"
function array[5](){ echo "you are inside the function"; }
array[7]="seventh"
echo ${!array[*]}

Only returns 0 7.

So, how can I add a function to an array element, so that I can loop over it?

gniourf_gniourf · Accepted Answer · 2014-11-11 16:56:46Z

6

You can't really put the function itself in the array. But what you can do is put the function name in that field. Look:

a_pretty_function() {
    echo "you're inside a function";
}
array[5]=a_pretty_function

# Execute it:
"${array[5]}"

When you're doing

array[5]() { echo "you are inside the function"; }

you're defining a function called array[5] as you can check with declare -pf array[5].

With this mechanism you can do something horribly ugly:

array[5]() { echo "you are inside the function"; }
i=5
"array[$i]"

and this will execute the function array[5]. I wouldn't recommend such a practice in any case.

edited Nov 11, 2014 at 16:56

answered Nov 11, 2014 at 16:47

gniourf_gniourf

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2 Comments

Etan Reisner Over a year ago

Indeed the key is that last bit. Variables and functions don't share a namespace.

chepner Over a year ago

It also demonstrates that array[5] isn't so much an index operation as a name that bash treats specially for parameter expansion. Arrays are just ugly in bash.

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