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On CodeReview I posted a working piece of code and asked for tips to improve it. One I got was to use a boolean method to check if an ArrayList had an even number of indices (which was required). This was the code that was suggested:
private static boolean isEven(int number) { return (number & 1) == 0; } As I've already pestered that particular user for a lot of help, I've decided it's time I pestered the SO community! I don't really understand how this works. The method is called and takes the size of the ArrayList as a parameter (i.e. ArrayList has ten elements, number = 10).
I know a single & runs the comparison of both number and 1, but I got lost after that.
The way I read it, it is saying return true if number == 0 and 1 == 0. I know the first isn't true and the latter obviously doesn't make sense. Could anybody help me out?
Edit: I should probably add that the code does work, in case anyone is wondering.
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Answer: Boolean is a primitive data type that takes either “true” or “false” values. So anything that returns the value “true' or “false” can be considered as a boolean example. Checking some conditions such as “a==b” or “a<b” or “a>b” can be considered as boolean examples.
Boolean values and operationsConstant true is 1 and constant false is 0. It is considered good practice, though, to write true and false in your program for boolean values rather than 1 and 0.
A Boolean value represents a truth value; that is, TRUE or FALSE. A Boolean expression or predicate can result in a value of unknown, which is represented by the null value.
To check if a value is of boolean type, check if the value is equal to false or equal to true , e.g. if (variable === true || variable === false) . Boolean values can only be true and false , so if either condition is met, the value has a type of boolean. Copied!
Keep in mind that "&" is a bitwise operation. You are probably aware of this, but it's not totally clear to me based on the way you posed the question.
That being said, the theoretical idea is that you have some int, which can be expressed in bits by some series of 1s and 0s. For example:
...10110110 In binary, because it is base 2, whenever the bitwise version of the number ends in 0, it is even, and when it ends in 1 it is odd.
Therefore, doing a bitwise & with 1 for the above is:
...10110110 & ...00000001 Of course, this is 0, so you can say that the original input was even.
Alternatively, consider an odd number. For example, add 1 to what we had above. Then
...10110111 & ...00000001 Is equal to 1, and is therefore, not equal to zero. Voila.
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You can determine the number either is even or odd by the last bit in its binary representation:
1 -> 00000000000000000000000000000001 (odd) 2 -> 00000000000000000000000000000010 (even) 3 -> 00000000000000000000000000000011 (odd) 4 -> 00000000000000000000000000000100 (even) 5 -> 00000000000000000000000000000101 (odd) 6 -> 00000000000000000000000000000110 (even) 7 -> 00000000000000000000000000000111 (odd) 8 -> 00000000000000000000000000001000 (even) & between two integers is bitwise AND operator:
0 & 0 = 0 0 & 1 = 0 1 & 0 = 0 1 & 1 = 1 So, if (number & 1) == 0 is true, this means number is even.
Let's assume that number == 6, then:
6 -> 00000000000000000000000000000110 (even) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 1 -> 00000000000000000000000000000001 ------------------------------------- 0 -> 00000000000000000000000000000000 and when number == 7:
7 -> 00000000000000000000000000000111 (odd) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 1 -> 00000000000000000000000000000001 ------------------------------------- 1 -> 00000000000000000000000000000001 If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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