解释下面的C ++方法 [英] Explain the following C++ method

问题描述

#define XL     33           
#define OR      113          
#define NOR     313     
#define TN     344  

int to_bits(int critn,char *mask)
{
       unsigned int x;
       int begin;

       if (critn < XL)           begin = 1;
       else if (critn < OR)      begin = XL;
       else if (critn < NOR)     begin = OR;
       else if (critn <= TN)    begin = NOR;
       else                        begin = 0;
       x = critn - begin;

       *mask = (char)(0x80 >> (x % 8));

       return (int)(x >> 3);    // fast divide by 8  
}

我不知道C ++代码。

I don't have any knowledge of C++ code. Can any one explain what this method is doing in the last 2 lines?

感谢

推荐答案

在C ++中，就像大多数编程语言一样，您只能返回一个值。要返回两个值，这是一个常见的C / C ++实践，返回一个并传递一个指针到一个对象，并通过指针修改该对象（ mask 在这种情况下）

In C++, just like most programming languages, you can only return one value. To "return" two values, it's a common C/C++ practice to return one and pass a pointer to an object and modify that object via the pointer (mask in this case).

mask 指向的对象将被分配一个只有一个位设置的位掩码。这是通过获取十六进制值0x80（二进制形式为1000 0000），然后右移0到7步。确切的步数由 x 决定，它是使用某些特定应用程序逻辑的计算机。

The object that mask point to will be assigned a bitmask with exactly one bit set. This is done be taking the hexadecimal value 0x80 (1000 0000 in binary form) and right shift it 0 to 7 steps. The exact number of steps is decided by x, which is computer using some application-specific logic.

返回的是 x / 8 。

你可以看到例程作为一个除法例程返回 x / 8 和余数（如 x modulo 8 ，但表示为位掩码而不是整数值）。

You can see the routine as a division routine that returns x/8 and the remainder (like x modulo 8, but expressed as a bit mask rather than an integer value).

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