Tom Lane wrote: > "Tony S" <[EMAIL PROTECTED]> 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. ---------------------------(end of broadcast)--------------------------- TIP 2: Don't 'kill -9' the postmaster
