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s5->s7 subtraction of LIR result

Created by: lambertws at: 12/10/2010 5:08 AM (2 Replies)

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12/10/2010 5:08 AM	Rate (0)
lambertws Posts: 58 Rating: (0)	I have code line like this and what I understand so far L KH E400 //absolute address of DB 0 L FY 232 //offset of the absolute address SLW 1 //times 2 because each DB have 2 wide absolute address +F LIR 0 //load the address of DW 0 of the DB pointed by previous line T FW 236 //transfer the address to FW 236 L KF 0 !=F //compare the address to 0?? is it for LIR error checking? JC M002 L FW 236 //load the address to accumulator L KF 2 -F //subtraction of DW 0 address?? point to what then?? DW of other DB? LIR 0 //the result of LIR is getting LIRed L KF 5 -F //subtraction of the LIR result once more L KF 127 <=F L KY 1,103 JC M002 DO FW 231 C DB 0 JU M003 anyone can help me?
lambertws Posts: 58 Rating: (0)
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12/11/2010 10:53 PM	Rate (0)
jklm Platinum Member Joined: 12/22/2006 Last visit: 1/25/2022 Posts: 1742 Rating: (303)	Hello, 1) ============ LIR 0 //load the address of DW 0 of the DB pointed by previous line T FW 236 //transfer the address to FW 236 L KF 0 !=F //compare the address to 0, testing if DB exists JC M002 This part is testing if the DB is available (0 means that there is no start address for the DB). 2) ======== L FW 236 //load the address to accumulator L KF 2 -F //subtraction of DW 0 address, points into the header of the DB LIR 0 //the result into ACCU1 L KF 5 -F //subtraction of the LIR result once more L KF 127 <=F I believe that this part is reading the DB length in words (header included) Because the header is 5 words long, substract header length to know the dataword count. Than test if the dataword count is less equal than 127.
	Best regards jklm Suggestion was helpful? Feel free to rate it.
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12/22/2010 2:17 AM	Rate (0)
lambertws Posts: 58 Rating: (0)	hello jklm, thanks for the response. I try to do it your way because I have no idea about that lines. I will give feed back once I get the result. regards.
lambertws Posts: 58 Rating: (0)
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