assembly - memory location calculation -


the question follow : microcomputer has memory locations 00000h fffffh. each memory location stores 1 byte. in decimal, how many bytes can microcomputer store in memory? how many kilobytes this?

answer: requires 2 2kbytes of ram , 512 bytes of prom.

i try calculate myself before reviewing answer,and find out not same,till still don't understand why answer,anyone may give me help??thanks

well, 00000h fffffh, represents 100,000h memory locations, hence 100,000h bytes. 10h^5 (hex) 16^5 (decimal).

16^5 = (2^4)^5 = 2^20 = (1024)^2 = 1 m = 1024 k.

conclusion: processor can address 1 megabyte of memory.

obviously, less 1 megabyte installed on microcomputer, , not of installed memory ram. can't deduce amount of addressable memory.


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