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Computing + ARM Questions

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  • 18-10-2015 10:56pm
    #1
    RoRo979
    Registered Users Posts: 476 ✭✭


    hi, I am stuck on whether my answers for these are right/wrong and am looking for guidance.

    1) How many unique values can be represented using a binary numeral system with: (i) 6 bits
    is this just 6^2 or is it 6^2 -1?

    2) If one hundred integers are stored consecutively in memory beginning at address 0x2000 (200016) and if each integer requires one word of storage, what will be the address of the word containing the 22nd integer?
    Right so i get that each integer will be 4 bytes, so the 22 integer will finish at 88 bytes in or 0x2088 and start at 0x2085? I remember hearing there is something to do with hexadecimal here.

    3) Assuming the ARM architecture definition of the word and halfword units of storage, calculate:
    (i) the number of unique values that can be stored in a halfword
    (ii) the range of non-negative integers that can be stored in a halfword
    (iii) the number of unique values that can be stored in a word
    (iv) the range of non-negative integers that can be stored in a word
    so is it:-
    i) 16^2
    ii)0-15
    iii) 32^2
    iv) 0-15

    and q7 attached as a photo which I am completely clueless on how to do.Thank you


Comments

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    #2
    ED E
    Registered Users Posts: 36,167 ✭✭✭✭ED E


    Hey boards, do my homework for me!


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    #3
    RoRo979
    Registered Users Posts: 476 ✭✭RoRo979


    ED E wrote: »
    do my homework for me!

    I have already put up the answers I think they are. These are concepts I want to grasp properly.


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    #4
    PrzemoF
    Registered Users Posts: 1,931 ✭✭✭PrzemoF


    1. How many unique values can be represented using a decimal numeral system with: (i) 1 digit?
    9 is the max number, but zero counts as well, so 10

    2. 22 x 4 is 88, but you're mixing nicely hex with dec. 0x58 = 88 in decimal 5 x 16 + 8 = 88

    3i and 3iii - it's the same question as 1. provided that you know the length of halfword/word

    3ii and 3 iv - number of unique values = range of negative values + 0 + range of positive values. One bit is for the sign..

    7 Writing assembler when you're not sure about binary/dec/hex stuff might be interesting..


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