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Convert the following from hex to binary and draw it on the memory map.
RAM = 0000 -> 00FF EPROM = FF00 -> FFFF
Answer:
0000 0000 0000 0000 (0) RAM start
0000 0000 1111 1111 (255 byte) RAM stop
1111 1111 0000 0000 (63.75K) Eprom start
1111 1111 1111 1111 (64K) Eprom stop The memory map is shown below:-
Due to the tri-state nature of the I.C's (Ram, Rom etc.) they may be placed in parallel and 'turned on ' when they are required. This process is termed decoding. There are two types of decoding:- 1. Partial decoding 2. Absolute decoding
Q. Polling for an I/O completion is able to waste a large number of CPU cycles if the processor iterates a busy-waiting loop several times before the I/O completes. However if the
how it is work.
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Usually memory is allocated from a large pool of unused memory area called the heap. In C++, dynamic allocation/deallocation must be manually performed using commands like malloc,
Additional Reference bits algorithm Here we keep an 8-bit byte for every page in memory. At standard intervals the reference bit is shifted by the OS. If a shift register conta
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Addventage
What is preallocation? Why do it? Allocating space for a file before making the file to allow for expansion. This reserves space for a certain file so that other files can't gr
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