My son asked me this question, if a computer has 32 bit wide memory word, a 32 bit address bus and the memory is word-addressable, HOW MANY 8 BIT BYTES OF DATA CAN BE HELD WITHIN THE MEMORY, if all the possible memory locations are used ? In KiloBytes (KB) MegaBytes (MB) or GigaBytes (GB)
2006-08-17 21:39:48 · 7 answers · asked by Raja Idris R 2 in Computers & Internet ➔ Hardware ➔ Desktops
If you have 32 bits of addresses, each one holding 4 bytes of data, you have 17,179,869,184 bytes of data. (2^32) * 4 = 17,179,869,184.
That is 17 GB of data. That doesn't mean the computer can accept 17GB of RAM, just that it has enough addresses to access that many locations.
2006-08-17 23:20:09 · answer #1 · answered by Anonymous · 1⤊ 0⤋
i guess you multiply (2 to the power of 32) by itself and divide by 8.
the number that brings up is ludicrisly big, so i wouldnt worry.
computers may offer 32 bit words in memory but they are probebly just using 4 bytes of normal memory.
good question though.
2006-08-17 21:50:50 · answer #2 · answered by Anonymous · 0⤊ 0⤋
1Byte = 8 bits
1bit is the lowest unit for measuring digital data.
So a DWORD(32bit wide memory word) has 4Bytes.
2006-08-17 22:02:52 · answer #3 · answered by Kaz 2 · 0⤊ 0⤋
Whether your son asked you or not, the answer should be 4
2006-08-17 21:58:00 · answer #4 · answered by Anonymous · 1⤊ 0⤋
your son asked you that, eh?
I'm not sure I believe that.
If I come up with an answer, I'll put it up here.
2006-08-17 21:47:09 · answer #5 · answered by Jim T 6 · 0⤊ 0⤋
It is just common sense!
2006-08-17 21:49:59 · answer #6 · answered by L 5 · 0⤊ 0⤋
Use you head.
2006-08-17 21:44:48 · answer #7 · answered by Anonymous · 0⤊ 0⤋