一个字节有 8 个位, 这些位可能是 0 也可能是 1; 现在要算出一个字节中是 1 的位共有多少个.

第一种方法是一个函数;
第二种方法笨了点, 是先把 256 种可能值给一个数组, 随时调取.

第一种方法虽然灵巧, 但不如第二种方法快(作者书中说: 在非特殊情况下, 一般要快到 10 倍左右);
第二种方法虽然快捷, 并且使用方便, 但要以 256 个字节的数组空间为代价.

unit Unit1;interfaceusesWindows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,Dialogs, StdCtrls;typeTForm1 = class(TForm)Button1: TButton;procedure Button1Click(Sender: TObject);end;varForm1: TForm1;implementation{$R *.dfm}{方法1: 获取函数}
function GetByteBits(x: Byte): Byte;
beginResult := 0;while x <> 0 dobeginif Odd(x) then Inc(Result);x := x shr 1;end;
end;{方法2: 把所有可能的值放在一个常数数组}
constBitArr: array[0..MAXBYTE] of Byte = (0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8);{测试}
procedure TForm1.Button1Click(Sender: TObject);
varb,num: Byte;
beginb := 255;num := GetByteBits(b);      {使用函数获取}ShowMessage(IntToStr(num)); {8}num := BitArr[b];           {直接使用数组获取}ShowMessage(IntToStr(num)); {8}b := 254;num := GetByteBits(b);      {使用函数获取}ShowMessage(IntToStr(num)); {7}num := BitArr[b];           {直接使用数组获取}ShowMessage(IntToStr(num)); {7}
end;end.

那个小函数, 琢磨了半天才明白(惭愧); 以后判断其他数也没问题了, 譬如判断 Integer:

function GetIntBits(x: Integer): Byte;
beginResult := 0;while x <> 0 dobeginif Odd(x) then Inc(Result);x := x shr 1;end;
end;