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批踢踢實業坊 Examination 板
[考題] 中華電信 100年計算機概論
Jul 4th 2013, 23:34, by TPEer

作者TPEer (Yuan)

看板Examination

標題[考題] 中華電信 100年計算機概論

時間Thu Jul 4 23:34:13 2013

一記憶體位址範圍為 6000(16)~8FFF(16),每一位址可儲存 16 位元,試問該記憶體 容量為多少 KB(Kilo Bytes)? (A)12 KB (B)24KB (C)48 KB (D)96 KB 小弟這題查到的算法解到 8FFF(16)-6000(16)=2FFF(16) 到這都還OK,但是下一步... 2FFF(16)+1=3000(16) 這地方有點不懂 這個3000是怎麼得來的呢? 懇請板上高手幫忙解惑一下... -- Follow my heart -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.45.227.200

p1235716666:0 1 2 3 4 5 6 7 8 9 總共幾個數? 07/04 23:43

p1235716666:9-0=9 然後還要再加1 07/04 23:44

TPEer:就是這裡算完困惑 2FFF(16)是怎麼解開然後+1去得到3000 07/04 23:46

a1988152002:F+1=16寫0進一 F+1=16寫0進一 F+1=16寫0進一 2+1=3 07/04 23:49

a1988152002:3000 07/04 23:49

NCKU:ex:0000(16)~0001(16) 你覺得是1個還是2個位址 07/04 23:53

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