Bit Setting for long long

This is a discussion on Bit Setting for long long within the C++ Programming forums, part of the General Programming Boards category; I'm trying to write an expression in a single line that can do the following: take the sizeof(type), then fill ...

  1. #1
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    Bit Setting for long long

    I'm trying to write an expression in a single line that can do the following:

    take the sizeof(type), then fill a long long variable with 1's for the amount of bytes given by the sizeof.

    For example:

    type = long - then my number is: 0x00000000ffffffff

    type = long long - then my number is 0xffffffffffffffff

    Thanks!

  2. #2
    C++ Witch laserlight's Avatar
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    How will you handle this type X?
    Code:
    struct X
    {
        long long a, b;
    };
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  3. #3
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    i need to do this:

    static const double SCALE = 2.0 / VALUE;

    where value is the variable i need to calculate with the sizeof. So I can fill half of VALUE with type long, and all of VALUE with type long long.

  4. #4
    i've lost my mind
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    edit: iMalc gave a way better example below, I removed my /fail pseudo code.

    The C++ witch makes a good point, not only could you receive a struct larger than any of the basic/standard data types, but you could receive a struct with two or three chars, you won't actually know what you're dealing with unless you require some extra input.

    I don't know what you're trying to do, but it may not work out like you want without convoluted functions (eww gross!) in this approach. ;p
    Last edited by gltiich; 10-13-2009 at 01:53 PM.

  5. #5
    The larch
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    May-be a template function:

    Code:
    #include <iostream>
    template <class N>
    N all_ones()
    {
        return ~N();
    }
    
    int main()
    {
        unsigned long long a = all_ones<unsigned long>();
        unsigned long long b = all_ones<unsigned long long>();
        std::cout << std::hex << a << '\n' << b << '\n';
    }
    This won't work with signed types, though, because the result would be seen as -1 in either case, and the implicit cast from -1L to -1LL would flip the rest of the bits.

    However, it also appears to me that you might just want to divide 2.0 by the maximum value of corresponding unsigned type:

    Code:
    #include <limits>
    
    double d = 2.0 / std::numeric_limits<unsigned long>::max();
    Be warned that you could lose some precision with unsigned long long, since its maximum value might not be exactly representable as double.
    Last edited by anon; 10-13-2009 at 08:59 AM.
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  6. #6
    Algorithm Dissector iMalc's Avatar
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    ~0ULL>>(64-sizeof(type)*8) should do it .
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