Tom Lane wrote:
> "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
This is very much my mistake. I had indeed taken them to be a sort of
pass-by-reference parameter, and not part of the result definition,
which they actually are. Running PERFORM is pointless, then, too.
