Hi, my assumption was that Bash's "printf" builtin implicitly defines a local variable when used inside a function like so:
function foobar { printf -v foo bar; }
foobar
declare -p foo
# Prints "bar" from the global "foo" variable.
But instead I have to declare the "foo" variable to be a "local" member of
the "foobar" function like so:
unset -v foo
function foobar { local foo=; printf -v foo bar; }
foobar
declare -p foo
# Prints an error message saying that there is no "foo" defined.
Cheers
Tim
