what does this mean, bitwise operator

This is a discussion on what does this mean, bitwise operator within the C++ Programming forums, part of the General Programming Boards category; im kinda confused on what exactly this means/does. Code: if (n & 1) //action n is a parameter to a ...

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    what does this mean, bitwise operator

    im kinda confused on what exactly this means/does.
    Code:
    if (n & 1)
    //action
    n is a parameter to a recursive function and the function works, but what does the & (bitwise operator) does. From what i can understand is that when n is eventually 1 then the recursive function is called for the last time and then it returns the result. Am i right?
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    Registered User hk_mp5kpdw's Avatar
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    Lets say that (in binary) the variable n has the following contents:

    01010101010101010101010101010101

    When we perform the bitwise AND of that value with 1 which is:

    00000000000000000000000000000001

    ... we get the resulting value:

    00000000000000000000000000000001

    ...which will evaluate to 'true' for the purposes of the if test. Basically the test is determining if the low order bit is set.
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
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    Slave MadCow257's Avatar
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    n is a parameter to a recursive function and the function works, but what does the & (bitwise operator) does. From what i can understand is that when n is eventually 1 then the recursive function is called for the last time and then it returns the result. Am i right?
    Yes, but it really only means that when n = 0 it will end

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    n & 1 evaluates to 1 if the last bit in the number is set, and it evaluates to 0 if it is not. Since 1 means true and 0 means false, the action is performed if the last bit is set in the number. If n is unsigned, then (n & 1) will be true whenever n is odd, so whether n is eventually 1 or 3 or some other odd number doesn't matter. It will end when n is even.

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    so that means that 0 is considered an even number?
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    Mad OnionKnight's Avatar
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    n&1 is also the same as n%2

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    Quote Originally Posted by InvariantLoop
    so that means that 0 is considered an even number?
    What is:

    0000 0000
    0000 0001 &
    ------------


    ?

    What will happen to the loop when n=0?
    Last edited by 7stud; 03-09-2006 at 05:07 PM.

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    >> so that means that 0 is considered an even number?
    Yes, 0 is even because of the definitions of even and odd. http://en.wikipedia.org/wiki/Even_and_odd_numbers

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    Quote Originally Posted by Daved
    >> so that means that 0 is considered an even number?
    Yes, 0 is even because of the definitions of even and odd. http://en.wikipedia.org/wiki/Even_and_odd_numbers
    So, if wikipedia had defined 0 as an odd number, would that affect what the loop does when n=0?
    Last edited by 7stud; 03-09-2006 at 05:48 PM.

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    No. Your question doesn't really make sense given the context of my statement. I think you may have misunderstood my point.

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    0 is defined mathematically as an even number. A formal mathematical definition is that an integer i is even if it is exactly equal to 2*j where j is also an integer. All integers that are not even are odd. This definition can be expressed in a number of equivalent forms (eg a number i is even if the remainder one dividing it by 2 is zero).

    Daved's link to wikipedia was for information, not a claim that life is defined by wikipedia.

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    An even number is a number that can be divided by 2 e.g. it doesn't leave any decimals/no remainder. Simple as that.

    0/2 = 0
    1/2 = 0.5
    2/2 = 1
    3/2 = 1.5
    4/2 = 2

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