Question

#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  
}

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

Thanks

Answer 1

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).

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.

The value returned is the x / 8.

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).