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To: Johan Van Houtven who wrote (5364)	7/12/1998 10:01:00 AM
From: Johan Van Houtven	Read Replies (1) \| Respond to of 11149

Bob, problem fixed. Thank you for your help! Looked at your parse-time error explanation again this morning after a good nights sleep and there it was: the perfect workaround! I just have to enclose the max() function with an if...endif pair as follows: if StartIndex <= 0 then Println symbol, ",", StartIndex, ",", Max(-1, StartIndex, hi); endif; Take away the if...endif and the scan presents you with a syntax error "Day value must be less <= 0". Even while the StartIndex IS a negative number. Sent the following bug report to Gary: --- end of email to Gary --- Please run the following scan: // start of scan input="test-stocks-in.lst"; output="test.lst"; integer i, Startindex; DaysToLoad=2200; StartIndex :=(DaysLoaded-1) * -1; Println symbol, ",", StartIndex, ",", Max(0, StartIndex, hi); // end of scan on this "test-stocks-in.lst" containing the following symbols: CATP EMC IM AMCC SCMM BEST You will note that you get an error message: "Syntax error: Day value must be less <= 0" The day value IS <= 0. You can easily check that in the following way: Change the Println line to this: Println symbol, ",", StartIndex; Run the scan and you'll get this: CATP ,-1326 EMC ,-1394 IM ,-424 AMCC ,-155 SCMM ,-190 BEST ,-195 All the StartIndex values are there, and they are all correctly negative numbers. So the "Syntax error: Day value must be less <= 0" is INCORRECT. Therefor the bug must be in the Max() parsing routine. I used a workaround presented to me by Bob Jagow that makes the scan work. You have to enclose the Prinlln statement with and if endif as follows: if StartIndex <= 0 then Println symbol, ",", StartIndex, ",", Max(-1, StartIndex, hi); endif; Thank you. --- end of email to Gary --- Awesome program QPv2 and you guys in the thread, especially Bob, are just as awesome. Great thread guys. Thx. Johan