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Load indirectly a register with a memory location
Created by: SG12345 at: 3/15/2010 9:40 AM (1 Replies)
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| 3/19/2010 1:59 PM | |
Joined: 10/7/2005 Last visit: 11/7/2024 Posts: 3026 Rating:
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Hello SG12345 You basically understood it correctly, but allowme to elaborate furtehr ona few important aspects: Address 7EAA will (in a 95U) contain the address of where DW0 of DB170 currently is located in the PLC's RAM. You can comparethe address area 7E00 to 7EFF (applicable for a 95U) to aFile Allocation Table (FAT) of a PC's hard disk. Documents can "move around" on the hard disk (if you change them or compress the HD) but theFAT is fixed andcontains a reference towhere the document in question currently resides on the HD. The Block Address List in your95U is alsoin a fixedpart of the RAM, whereas the Blocks themselves can "move around"in the RAM(if you change, download orcompress the PLC). By reading the value of 7EAAyou will however find out where DB170 currently starts in the RAM (irrespective of how many DB's there are in the PLC). What you do with that DB170 start address information in 7EAA is now entirely up to you (or whatever is contained in the rest of your existing program). Last but not least a word of caution: Direct memory addressing was a very powerful featureof S5's. It canhowever fraud with danger as it bypasses all safety checks that "normal" addressing has. A simple typo or incorrect calculation can for example lead toaddressing a memory location where adifferent DB - or worse -OB/PB/SB/FB/FX currently resides. Worst case scenario is absolute memory corruption with an overall reset being the onlyway out of it (I don't mean to scare you, but you ought to have respect for using this kind of addressing and would be well advised to back you program up before you do your changes). I hope this helps |
Last edited by: fritz at: 3/20/2010 2:38 AMFixed typos Cheers |
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