右移2执行鸿沟在-1 [英] Right Shift to Perform Divide by 2 On -1

问题描述

我知道我可以通过2使用右移执行鸿沟。

I know that I can perform divide by 2 using right shift.

为了简单起见，取4位数字系统

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

如果我试图执行

6 / 2 = 0110 >> 1 = 0011 = 3
-6/ 2 = 1010 >> 1 = 1101 = -3

都有效+ ve和-ve号

Is valid for both +ve and -ve number

然而，当走到1

1 / 2 = 0001 >> 1 = 0000 = 0
-1/ 2 = 1111 >> 1 = 1111 = -1

好像有在-1一个特殊的情况下，右移，然后将其移动到负无穷大。

Seems like there is a special case in -1, as right shift then to move it to negative infinity.

目前，我需要把一个特殊的，如果检查这一点，因为我期待-1 / 2 = 0。

Currently, I need to put a special if check for this, as I am expecting -1 / 2 = 0.

我想知道你是怎么家伙处理此异常在code？你小子放一个，如果检查？

I was wondering how do you guy handle this exception in your code? You guy put an if check?

推荐答案

任何负面的奇数将无法正常工作。然而，为了回答你的问题，如果你知道你可以有负数，只是2.本变成了修正转向由JIT /编译器划分。

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.

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