Right Shift to Perform Divide by 2 On
I know that I can perform divide by 2 using right shift.
For simplicity, take a 4 bit number system
-1 - 1111
-2 - 1110
-3 - 1101
-4 - 1100
-5 - 1011
-6 - 1010
-7 - 1001
-8 - 1000
7 - 0111
6 - 0110
5 - 0101
4 - 0100
3 - 0011
2 - 0010
1 - 0001
0 - 0000
If I try to perform
6 / 2 = 0110 >> 1 = 0011 = 3
-6/ 2 = 1010 >> 1 = 1101 = -3
Is valid for both +ve and -ve number
However, when come to 1
1 / 2 = 0001 >> 1 = 0000 = 0
-1/ 2 = 1111 >> 1 = 1111 = -1
Seems like there is a special case in -1, as right shift then to move it to negative infinity.
Currently, I need to put a special if check for this, as I am expecting -1 / 2 = 0.
I was wondering how do you guy handle this exception in your code? You guy put an if check?
Any negative odd number won't work. However to answer your question, if you know you can have negative numbers, just divide by 2. This is turned into a shift with a fixup by the jit/compiler.
@Anon is technically correct.
However, it is best practice to use the / operator for division, and leave micro-optimization to the JIT compiler. The JIT compiler is capable of optimizing divisions by constants as shift/add sequences ... when this is an optimal thing to do for the execution platform.
Doing this kind of thing is (probably) a premature optimization, and it may be an anti-optimization if your code needs to run fast on multiple Java platforms.
If you right-shift to divide by two, you always end up "rounding" down - towards zero if positive, away from it if negative.
If this is not what you want, you can correct for it:
if (n & 1 > 0 && n < 0)
result += 1;
上一篇: 以C ++为基础的本地化文本最佳实践
下一篇: 右移2进行分割