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Convert the following from hex to binary and draw it on the memory map.
RAM = 0000 -> 00FF EPROM = FF00 -> FFFF
Answer:
0000 0000 0000 0000 (0) RAM start
0000 0000 1111 1111 (255 byte) RAM stop
1111 1111 0000 0000 (63.75K) Eprom start
1111 1111 1111 1111 (64K) Eprom stop The memory map is shown below:-
Due to the tri-state nature of the I.C's (Ram, Rom etc.) they may be placed in parallel and 'turned on ' when they are required. This process is termed decoding. There are two types of decoding:- 1. Partial decoding 2. Absolute decoding
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Determine the syntax of the assembler directive EQU The following is syntax of the assembler directive EQU: EQU
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explain about oprating system services and topics briefly
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