Checkpoint 1.1:
（1）1The addressing capacity of a CPU is 8KB, so the width of its address bus is13position
Because 8KB=8*1024B=2^13
（2）1KBThe memory has1024The number of storage units and the number of storage units.0reach1023
1KB=1024B，Numbering from 2^0 to 2^10-1
（3）1KB=1024B=1024*8bit
（4）1KB=2^10B 1MB=2^20B 1GB=2^30B
（5）8080,8088,80286,80386The width of the address bus is 16, 20, 24, and 32, and the data they can transmit at one time are:64KB,1MB,16MB,4GB
If a CPU has N address lines, you can say that the width of the CPU address bus is N, and you can find 2 ^ N memory units.
（6）8080,8088,8086,80286,80386Data bus width bits 8, 8, 16, 16, 32, then they can transmit data at one time:1B,1B,2B,2B,4B
8The root data bus can transmit 8 bit binary at a time, that is, one byte.
（7）Read 1024 bytes of data from memory. 8086 read at least.512Second, 80386 at least read.256second
8086One can transmit 2B, 80386 can transmit 4B at a time.
（8）In memory, data and programs areBinaryForm storage