在1024x8内存芯片中寻址每个机器位置需要多少个地址线？ [英] How many address lines are needed to address each machine location in a 1024x8 Memory chip?

问题描述

请告诉我解决数据总线和地址总线相关问题的程序。

please tell me the procedure to solve data bus and address bus related questions.

推荐答案

这似乎是一项功课，我们不会为你做家庭作业。

只需提示访问1024个单元格（每个8位宽），您需要多个地址线，其二进制数最多为1024（考虑地址线为二进制数字）。

如果你在这个建议之后无法解决，也许你想重新考虑一下，如果IT是你正确的学习领域...

This seems an homework, and we don't do homeworks for you.
Just a hint to access 1024 cells (each 8 bits wide) you need a number of address lines that counts up to 1024 in binary (consider an address line a binary digit).
If you can't solve after this suggestion maybe you want reconsider if IT is your right study field...

1024 x 8表示内存位置数= 1024。地址线= n然后2 ^ n =位置......就在这里2 ^ n = 1024 ..我们必须找到n？ n = 10 ..所以地址行的数量是10.这是正确的先生吗？

1024 x 8 means number of memory location = 1024. if number of address lines =n then 2^n = location... that is here 2^n = 1024 ..we have to find n ? n = 10 .. so number of address lines is 10. is it correct sir ?

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