Why doesn't dttofmtasc() ESQL functions work?

I am trying to use the ESQL function, dttoftmasc provided with Informix ESQL. I am running the following version of ESQL: INFORMIX-ESQL Version 5.01.UD1. The problem is that when I try to use a format string, I get an error -1228 (which is undefined). If I use a blank format string, the function returns 0 and returns the default format. Also, setting the environment variable DBTIME to anything has no effect on the format. Am I misusing this function or is this an Informix bug that I don't know about. If it is a known bug, how do I get around it? Below is the code and the output from running the code. Thanks in advance, TJ -- TJ Mills IV *---------------------------------------+--------------------------------------* | mills@SPK.HP.COM | PHYSICAL: T.J. Mills | AT&T: (509) 921-3661 | Hewlett Packard Co | HPDESK: Terence Mills/HP1000 | E 24001 Mission | TELNET: 1-921-3661 | Liberty Lake WA 99019 *---------------------------------------+--------------------------------------* EXEC SQL include datetime.h; EXEC SQL include decimal.h; #include <iostream.h> #include <Time.h> #include "/usr/local/mdm/include/Informix.h" main() { EXEC SQL BEGIN DECLARE SECTION; datetime YEAR TO MINUTE dt; datetime YEAR TO MINUTE dt2; EXEC SQL END DECLARE SECTION; char obuff[50], obuff2[50]; Time tj = Time(1988, Time::november,8); // init obuffs strcpy(obuff, ""); strcpy(obuff2, ""); // Init dt dtcurrent(&dt); dtcurrent(&dt2); // Store dt into obuff as ASCII int x = dttofmtasc(&dt, obuff, sizeof(obuff), "%m/%d/%Y %H:%M"); int y = dttoasc(&dt2, obuff2); // Print obuffs result cout << "Result = " << x << " Obuff = " << obuff << endl; cout << "Result = " << y << " Obuff2 = " << obuff2 << endl; // Store obuff into tj tj = make_time(obuff); // Print tj cout << tj << endl; } --------------------------What follows is the output ------------------------ Result = -1228 Obuff = Result = 0 Obuff2 = 1995-03-06 14:14 03/06/95 14:14:46