memory consists of many millionns of storage cell. Each cell can store a bit. data is accrssed in n-bit groups called word n is called word length
- 32 bit word len eg:
img : 32bit day6
b31 is a sign repr in bit
char = 1byte = 8bits :: therfore 4 chars can be stored
- num = one word
access by word address
- char = one byte
accessed by byte addr
- string = variable len
beginnning of the string = beginning byte addr
termination = special charr
- to retrieve from memory either for one word or byte sddr of each loc is needed a k-bit addr memory has 2 ^ k memory loc called memory space
24 bit= 16Mb
32 bit = 4Gb
1Kb = 2^10 =10bit
1Tb = 2^40
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it is impractical to assign addr to individual bit loc so only addr to success byte location - byte addressable memory
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word alignment - word are said to be alligned in memory if they begin at a byte addr that is a multiple of the number of bytes in a word eg:
- 8 bit word = 0,1,2,... locs
- 16 bit word = 0,2,4,... locs
- 32 bit word = 0,4,8,... locs
big-ending: lower byte addr are used for the most signigicant byte of the word
little-ending:oppposite ordering, lower byte addr are used for the least significant bytes of the word
prob:
- a memory has 32 bit addr and byte addresable what is the size of the memory(in byte)?
ans :2^32
2.a mem has 24 bit and word addresable with the word length of 32 bits, what si the size of the memory(in bytes)?
ans : (2^24 word)*2^2 (1 word = 4 bytes)
3.a mem has 16-bit addr and byte addressable word length is 32 bits(4bytes). How many word can we store in such a memory?
ans : (2^16 word)/2^2 (1 word = 4 bytes)