unset does not remove functions like a[b] unless -f is specified

[Emanuele Torre]

The unset builtin, when invoked without an option, should first try to
unset the variable (or array element) specified by its arguments, and
then fall back to trying to remove the function definition for the
function that has the name specified by the argument if it exists.

    bash-5.1$ declare -f a; declare -p a
    bash: declare: a: not found
    bash-5.1$ a=hi; declare -f a; declare -p a
    declare -- a="hi"
    bash-5.1$ unset a; declare -f a; declare -p a
    bash: declare: a: not found

    bash-5.1$ declare -f a; declare -p a
    bash: declare: a: not found
    bash-5.1$ a () { echo ;}; a=hi; declare -f a; declare -p a
    a () 
    { 
        echo
    }
    declare -- a="hi"
    bash-5.1$ unset a; declare -f a; declare -p a
    bash: declare: a: not found
    a () 
    { 
        echo
    }

    bash-5.1$ declare -f a; declare -p a
    bash: declare: a: not found
    bash-5.1$ a () { echo ;}; declare -f a; declare -p a
    a () 
    { 
        echo
    }
    bash: declare: a: not found
    bash-5.1$ unset a; declare -f a; declare -p a
    bash: declare: a: not found

The  -v  option can be used to tell the unset builtin to only unset
variables (and array elements), and not fall back to removing function
definitions.

The  -f  option can be used to tell the unset builtin to only remove
function definitions directly.

It seems that, if an operand of unset is in the form of
validvariablename[something], the unset builtin invoked without options,
will only attempt to delete an array element specified by that argument,
and will not fall back to removing a function named like the argument
(as if the -v option was passed only for that argument).
You must specify the  -f  option to delete a function with a name in
that form.

    bash-5.1$ a=([1]=2 3 4); a[] () { echo ;}; a[0] () { : ;}
    bash-5.1$ declare -p a; declare -f 'a[]' 'a[0]'
    declare -a a=([1]="2" [2]="3" [3]="4")
    a[] () 
    { 
        echo
    }
    a[0] () 
    { 
        :
    }
    bash-5.1$ unset 'a[]' 'a[0]' 'a[1]'
    bash-5.1$ declare -p a; declare -f 'a[]' 'a[0]'
    declare -a a=([2]="3" [3]="4")
    a[0] () 
    { 
        :
    }
    bash-5.1$ unset -f 'a[0]'
    bash-5.1$ declare -p a; declare -f 'a[]' 'a[0]'
    declare -a a=([2]="3" [3]="4")

Same result if the `a' variable does not exist.

    bash-5.1$ a[0] () { echo;}
    bash-5.1$ declare -p a; declare -f 'a[0]'
    bash: declare: a: not found
    a[0] () 
    { 
        echo
    }
    bash-5.1$ unset 'a[0]'
    bash-5.1$ declare -p a; declare -f 'a[0]'
    bash: declare: a: not found
    a[0] () 
    { 
        echo
    }
    bash-5.1$ unset -f 'a[0]'
    bash: declare: a: not found

I think that this is a bug, and that bash's unset builtin should fall
back to trying to remove the function named "a[0]" in those cases.

Cheers.
 emanuele6