Thread: Bits and Addresses

  1. #1
    Registered User
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    Jul 2007

    Bits and Addresses

    I'm writing a memory management program/simulation. Let's say I have a 32-bit pointer. I want it to represent a virtual address. Part of that address, let's say the upper 22 is the virtual page number, and the lower 10 is the page offset. How do I do bit selection in C++? How do I say something like:
    int upper = upper22bits(pointer);
    int lower = lower 10bits(pointer);


    I seem to remember messing around with shifts and masking but I don't really remember how.
    For example if I want the upper 22 its, I think I'd right shift 10, & with enough 1's to cover the bits (2^22-1 I think) and I'd have a number represented by the upper 22 bits.
    If I want the lower 10 bits I think I don't have to shift, I just & with enough 1's to cover the bits (2^10-1 I think).
    I'm not sure if this is correct

    I tested it out and it prints out the wrong stuff
    cout << "Testing bit selection" << endl;
    int test = 23; // 23 = "10111"
    cout << "Lower three bits: " << (test & 2^3-1) << endl; // "111" = 7
    cout << "Upper three bits: " << (test>>2 & 2^3-1) << endl; //"101" = 5
    Last edited by jcafaro10; 03-15-2009 at 02:15 PM.

  2. #2
    int x = *((int *) NULL); Cactus_Hugger's Avatar
    Join Date
    Jul 2003
    Banks of the River Styx
    2^3-1 is 0. However, the & will happen first (it has higher precedence that ^, see your Order of Operations)
    The bigger issue, however, I suspect that ^ is not what you're expecting it to be. ^ is binary XOR, not exponent. (There is no C operator for exponent.)
    long time; /* know C? */
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  3. #3
    Hurry Slowly vart's Avatar
    Join Date
    Oct 2006
    Rishon LeZion, Israel
    (There is no C operator for exponent.)
    actually when you need power of 2 you can use C operator - shift

    All problems in computer science can be solved by another level of indirection,
    except for the problem of too many layers of indirection.
    David J. Wheeler

  4. #4
    Registered User
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    Jul 2007
    Thanks, that did it

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