"Tony S" <tony@vectorsalad.com> writes:
> Function defined with INOUT parameter. Value of parameter is not returned
> to calling function.
You are confused about the meaning and use of INOUT. It's not some kind
of pass-by-reference parameter, it's just a shorthand for separate IN
and OUT parameters. In your example, the PERFORM discards the function
result; the original value of 'outparameter' is not and cannot be
modified by the called function.
regards, tom lane
