作者a124136906 (小麦)
看板Grad-ProbAsk
标题[理工] [计组]-中央94-电机
时间Wed Aug 12 16:51:48 2009
(1)A PC has 4 MB of RAM beginning at address 00000000H.
Calculate the last address(in hex) of this 4 MB block
(2)If the address and the ending address of the ROM block are 008000H and
010000H,calculate the size of the ROM in K.
Ans:(1)大小 = 4 Byte = 2的22次方 Byte
终止 = 大小 +起始 -1 = 400000的16进位 + 0 -1 =3FFFFF的16进位
(2)大小 = 终止 -起始 +1 = 010000H-008000H + 1 =008001H
=8 *16的3次方 + 1
=32K+ 1 Byte
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我想问第一提的400000的16进位怎麽来的!?!?!?想不出来.......
第二题的010000H-008000H + 1 为甚麽会等於008001H呢!?!?!?!?
希望各位大大可以帮我详解一下^^
因为我底子不是很强^^"
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※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 115.43.12.178
1F：推 sean456:2^22次方 变成HEX就是4000000 08/12 16:59
2F：推 sean456:第二题借位 变成16-8＝8 8000 08/12 17:03
3F：→ a124136906:谢谢楼上大大~想出来了!!感恩!! 08/12 17:13