# right shifting?

posted 5 years ago

So if low is 1 and high is 6 (110 base 2), the resulting value is 3, which is the median value if we had 6 numbers in range.

Source, Median - http://en.wikipedia.org/wiki/Median.

Luigi Plinge wrote:It's shifting all the bits to the right by 1. If you think about it, this is equivalent to dividing by 2 (and discarding the remainder). Shifting by 2 bits would be equivalent to dividing by 4 etc. So the statement just takes the average of low and high.

So if low is 1 and high is 6 (110 base 2), the resulting value is 3, which is the median value if we had 6 numbers in range.

Source, Median - http://en.wikipedia.org/wiki/Median.

Jon

Campbell Ritchie

Sheriff

Posts: 48910

58

posted 5 years ago

- 1

Luigi explained that shifting right one bit is the same as dividing by 2.

Long ago, this was a common trick to quickly divide a number by 2. The division instruction / on old processors was a relatively slow operation, while shifting one bit to the right was a much faster operation; so writing something like

So, I'd advise you to not use tricks like this in your own code; just write

Long ago, this was a common trick to quickly divide a number by 2. The division instruction / on old processors was a relatively slow operation, while shifting one bit to the right was a much faster operation; so writing something like

`number / 2`was significantly slower than writing`number >>> 1`. However, this isn't necessarily true anymore on modern microprocessors; and even if it were, then you should let the compiler make micro-optimizations like this.So, I'd advise you to not use tricks like this in your own code; just write

`number / 2`when you have to divide a number by two, because it's much clearer, and the compiler will be smart enough to optimize it if necessary.