如何在64位代码中实现高效的32位DivMod

对于总是除以10(每条评论)的特殊情况,您可以执行以下操作:

procedure DivMod10(num : Cardinal; var q, r : Cardinal); inline;
var
  rl : uInt64;
begin
  rl := UInt64(3435973837)*num;
  q := rl shr 35;
  r := num - q*10;
end;

算法因分母而异,但确定它的来源和幻数可以在libdivide中找到.这对于所有无符号32位整数都是准确的,并且比使用div(并提供余数)快约3倍.

基准(优化):

  t0 := GetTickCount;
  for I := 1 to 999999999 do begin
    DivMod10(i, q, r);
  end;
  ShowMessage(IntToStr(GetTickCount - t0));  // result :  1809

  t0 := GetTickCount;
  for I := 1 to 999999999 do begin
    q := i div 10;
  end;
  ShowMessage(IntToStr(GetTickCount - t0));  // result :  5336

测试:

for I := 1 to High(Cardinal) do begin
  DivMod10(i,q,r);
  if q <> (i div 10) then WriteLn(IntToStr(i));
  // no mismatch found
end;