中国DOS联盟

// 

// 计算参数x的平方根的倒数 

// 

float InvSqrt (float x) 

{ 

float xhalf = 0.5f*x; 

int i = *(int*)&x; 

i = 0x5f3759df - (i >> 1); // 计算第一个近似根 

x = *(float*)&i; 

x = x*(1.5f - xhalf*x*x); // 牛顿迭代法 

return x; 

} 

In the Quake3 engine, there is such a "mars algorithm":

// 

// Calculate the reciprocal of the square root of parameter x 

// 

float InvSqrt (float x) 

{ 

float xhalf = 0.5f*x; 

int i = *(int*)&x; 

i = 0x5f3759df - (i >> 1); // Calculate the first approximate root 

x = *(float*)&i; 

x = x*(1.5f - xhalf*x*x); // Newton's iteration method 

return x; 

} 

Personally, I think that set /a supports bit operations, and it should be possible to implement the reciprocal of the square root.

在 SET 命令中添加了两个新命令行开关:



    SET /A expression

    SET /P variable=



/A 命令行开关指定等号右边的字符串为被评估的数字表达式。该表达式

评估器很简单并以递减的优先权顺序支持下列操作:



    ()                  - 分组

    ! ~ -               - 一元运算符

    * / %               - 算数运算符

    + -                 - 算数运算符

    << >>               - 逻辑移位

                       - 按位“与”

    ^                   - 按位“异”

    |                   - 按位“或”

    = *= /= %= += -=    - 赋值

      &= ^= |= <<= >>=

    ,                   - 表达式分隔符

```
In the SET command, two new command-line switches are added:

SET /A expression
SET /P variable=

The /A command-line switch specifies that the string to the right of the equal sign is a numerical expression to be evaluated. This expression evaluator is simple and supports the following operations in decreasing order of precedence:

() - Grouping
! ~ - - Unary operators
* / % - Arithmetic operators
+ - - Arithmetic operators
<< >> - Logical shifts
- Bitwise "AND"
^ - Bitwise "XOR"
| - Bitwise "OR"
= *= /= %= += -= - Assignment
&= ^= |= <<= >>=
, - Expression separator
```
　　Microsoft's explanations have always been vague. Except for being able to use arithmetic operators and assignment symbols proficiently, I have never understood the rest. Please explain if anyone has used it.

　　In CMD, you can't directly calculate decimal multiplication; you still have to operate bit by bit. If you can understand how the set /a symbols are used, the problem of square roots is estimated to be solved easily.

Square root can be solved using Pascal's triangle

Using set /a is too cumbersome, and division is anyway cumbersome

Last edited by he200377 on 2006-10-16 at 00:34 ]