推 sean456:2^22次方 變成HEX就是4000000 08/12 16:59
推 sean456:第二題借位 變成16-8＝8 8000 08/12 17:03
→ a124136906:謝謝樓上大大~想出來了!!感恩!! 08/12 17:13
(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
終止 = 大小 +起始 -1 = 400000的16進位 + 0 -1 =3FFFFF的16進位
(2)大小 = 終止 -起始 +1 = 010000H-008000H + 1 =008001H
=8 *16的3次方 + 1
=32K+ 1 Byte
-----------------------------------
我想問第一提的400000的16進位怎麼來的!?!?!?想不出來.......
第二題的010000H-008000H + 1 為甚麼會等於008001H呢!?!?!?!?
希望各位大大可以幫我詳解一下^^
因為我底子不是很強^^"
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 115.43.12.178