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Patrick.Parks	Bug in Code Walker? Posted: Tuesday, July 25, 2006 2:25 PM (EST) Hi Paul, I agree with your assessment. F10 should stop on line [2] in both cases WITHOUT signalling the SYNTAX ERROR. This is a bug. However, I do NOT think this is happening because we are trying to EXECUTE line [2] without stopping after F10. It is probably happening beause we are COMPILING line [2] proactively and discovering the SYNTAX ERROR. Pat
Mark.Osborne	Bug in Code Walker? Posted: Tuesday, July 25, 2006 3:50 PM (EST) Paul,   I think Pat's assessment is on the mark.  Furthemore, F10 requests a step to line [2].  The interpreter needs to hang it's hat somewhere on the beginning of the line[2] codestring, but there isn't one yet, so there is no place for the token pointer to point to when stopping at the beginning of line[2].  Before we can stop there, we need to create the codestring by parsing the line.  That's when we get the syntax error.  Given the syntax error, there is no way to build a correct codestring to stop on.  So we have no alternative but to signal the error and suspend.   I don't think there is much else we can do here in any trivial way.   Am I missing something here and this is somehow more important to you than it appears to me?   /Mark
Paul.Ravitz	Bug in Code Walker? Posted: Tuesday, July 25, 2006 4:03 PM (EST) It's not terribly important. The example I gave was a simplification of an actual occurrence that happened to me today, to make it easier for others to duplicate it. It surprised me (that it appeared to execute), so I thought it was worth reporting. But since it's not actually executing but is only in an analysis stage, no harm is being done, so I don't think it should be high on anyones list. However, can you explain why it IS able to stop (apparently before creating the codestring) with F10 when there's a STOP on the line? Is it that the stop gives it a place "to hang it's hat"? Thanks. p
Mark.Osborne	Bug in Code Walker? Posted: Tuesday, July 25, 2006 4:21 PM (EST) Paul,   The STOP bit is stored in a byte in a codestring.  So setting a STOP on a line that does not yet have a codestring causes a dummy codestring to be materialized for that line.  It doesn't do syntax analysis, just drops a canonical dummy codestring in as a storage point for the stop bit.  So you don't get the syntax error then until you try to step past the dummy codestring and cause the line to be parsed.   /Mark