Set with 2 k * 8 of 16 k * 8 bits of storage memory chips (address unit 0000 h ~ 3 FFFH, each chip address space continuous), the unit address 0 b1fh chip's smallest address number is:
A. 0000 h
B. 2800 h
C. 2000 h
D. 0800 h
The answer is A, but I think 0 b1fh in the second chip, it should be 0800 h!!
O why is A great god taught? I seriously doubt the answer wrong!

CodePudding user response:

A 2 k x 8 bit memory chips into 16 k x 8 bits of storage capacity of 16 KB, so need eight pieces of memory chips,

The address unit for 000 h ~ 3 FFFH storage gives, on average, eight,


So the first block address units for 000 h ~ 7 FFH, so on and address unit respectively:

0000 h ~ 7 FFH, 0800 h ~ 0 FFFH,

1000 h ~ 17 FFH, 1800 h ~ 1 FFFH,

2000 h ~ 27 FFH, 2800 h ~ 2 FFFH,

3000 h ~ 37 FFH, 3800 h ~ 3 FFFH,

Can be seen easily 0 b1fh in 0800 h ~ 0 BFFF interval period,

The answer is wrong