c++ bitwise operators with 64 bit integers? possible?

This is a discussion on c++ bitwise operators with 64 bit integers? possible? within the C++ Programming forums, part of the General Programming Boards category; my compiler keeps giving me errors when I try to left shift (<<) past 32 bits in what the visual ...

  1. #1
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    c++ bitwise operators with 64 bit integers? possible?

    my compiler keeps giving me errors when I try to left shift (<<) past 32 bits in what the visual c++ compiler is telling me is a 64 bit integer (type long long). Is there any way I can do this, or do I need to take an alternate approach?

  2. #2
    and the hat of wrongness Salem's Avatar
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    Something more descriptive than "giving me errors" would be useful, like for example actual code and actual error messages.

    This for example, works fine for me.
    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    int main ( ) {
        __int64 i = 1;
        i <<= 33;
        printf("&#37;I64x\n", i );
        return 0;
    }
    
    $ cl /nologo foo.c
    foo.c
    
    $ foo.exe
    200000000
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  3. #3
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    Not sure what you're trying to do, but this works for me:
    Code:
    #include <iostream>
    #include <iomanip>
    
    int main() {
        __int64 x = 1;
    	int i;
    	for(i = 0; i < 64; i += 5) {
    		std::cout << i << " " << std::hex << (x << i) << std::endl;
    	}
    }
    
    Output:
    0 1
    5 20
    10 400
    15 8000
    20 100000
    25 2000000
    30 40000000
    35 800000000
    40 10000000000
    45 200000000000
    50 4000000000000
    55 80000000000000
    60 1000000000000000
    [Edit: beaten to it by Salem]

    --
    Mats

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    ah I just realized my error, I was doing this:

    Code:
    #include <iostream>
    
    using namespace std;
    
    void main() {
    	__int64 a (0);
    	a ^= 1<<33;
    	cout << a << endl;
    }
    instead of this:

    Code:
    #include <iostream>
    
    using namespace std;
    
    void main() {
    	__int64 a (1), shift (1);
    	a ^= shift<<33;
    	cout << a << endl;
    }

    it was giving me an undefined behavior error, but now I realize the 'a' variable wasn't the problem, it was the constant '1<<33'. Thanks for the help!

  5. #5
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    Quote Originally Posted by chrisd1100 View Post
    ah I just realized my error, I was doing this:

    Code:
    #include <iostream>
    
    using namespace std;
    
    void main() {
    	__int64 a (0);
    	a ^= 1<<33;
    	cout << a << endl;
    }
    instead of this:

    Code:
    #include <iostream>
    
    using namespace std;
    
    void main() {
    	__int64 a (1), shift (1);
    	a ^= shift<<33;
    	cout << a << endl;
    }

    it was giving me an undefined behavior error, but now I realize the 'a' variable wasn't the problem, it was the constant '1<<33'. Thanks for the help!
    You should be able to do
    Code:
    a ^= 1I64 << 33;
    too.

    --
    Mats

  6. #6
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    thanks Mats

  7. #7
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    Sorry to hijack a finished thread, but what range of operations do 32-bit chips support on 64-bit integers?
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  8. #8
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    Quote Originally Posted by CrazyNorman View Post
    Sorry to hijack a finished thread, but what range of operations do 32-bit chips support on 64-bit integers?
    Most 32-bit chips don't support 64-bit integers natively, but there are ways to support 64-bit operations as multiple operations, and some special operations to "extend to 64-bit", such as divide of two 64-bit numbers into a 32-bit number, and a multiply of two 32-bit that becomes a 64-bit number. Most recent compilers support 64-bit for all binary and unary operators +, -, *, /, %, &, |, ^, ~ and !.

    It is really up to the compiler to fill in the blanks here. You can really build "infinite" integers using the range of instructions supported by the processor. A 128-bit addition would look like this:
    Code:
        mov eax,[a]       // low 32-bit
        add [b], eax
        mov eax, [a+4]   // Next 32-bit 
        adc [b+4], eax
        mov eax, [a+8]   // Next 32-bit 
        adc [b+8], eax
        mov eax, [a+12]   // Next 32-bit 
        adc [b+12], eax
    The instruction adc is "add with carry", where the carry-over from the previous calculation is carried from one calculation to the next - just like if you add 8 + 4, you get 2 and 1 in "carry", then add the carry to "nothing" gives 12.

    A similar principle can be used for subtract and shift operatrons. Logical operations don't even need a carry. Multiply and divide are a bit tricky, but far from impossible (it just takes a lot longer than doing it in hardware if it's not a "simple" operation).

    --
    Mats

  9. #9
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    It is really up to the compiler to fill in the blanks here.
    Right! In the old days of DOS & 16-bit Windows, running on 16-bit x86 processors, you could run C/C++ which requires 32 bits for a type long. There are 8-bit machines that run C/C++.

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