随笔 - 2146  文章 - 97 评论 - 11767 trackbacks - 253

最近发现一些代码, 甚至有一些专家代码, 在遍历数组时所用的数组长度竟然是 SizeOf(arr); 这不合适!

如果是一维数组、且元素大小是一个字节, 这样用看不出错误, 譬如:
var
  arr1: array[0..9] of Char;
  arr2: array[0..9] of Byte;
begin
  ShowMessageFmt('%d,%d,%d,%d',[Length(arr1), SizeOf(arr1),
                                Length(arr2), SizeOf(arr2)]);
  {显示结果: 10,10,10,10}
end;

但如果数组元素多于一个字节、或是多维数组的情况下, 就不行了, 举例:
const
  arr1: array[0..9] of Integer = (1,2,3,4,5,6,7,8,9,10);
  arr2: array[0..1, 0..3] of Integer = ((1,2,3,4), (5,6,7,8));
var
  arr3: array[Boolean] of Integer;
  arr4: array[Byte] of Integer;
begin
  ShowMessage(IntToStr(Length(arr1)));    {10}
  ShowMessage(IntToStr(SizeOf(arr1)));    {40}

  ShowMessage(IntToStr(Length(arr2)));    {2}
  ShowMessage(IntToStr(Length(arr2[0]))); {4}
  ShowMessage(IntToStr(Length(arr2[1]))); {4}
  ShowMessage(IntToStr(SizeOf(arr2)));    {32}

  ShowMessage(IntToStr(Length(arr3)));    {2}
  ShowMessage(IntToStr(SizeOf(arr3)));    {8}

  ShowMessage(IntToStr(Length(arr4)));    {256}
  ShowMessage(IntToStr(SizeOf(arr4)));    {1024}
end;

我们倒是可以利用这个原理, 迅速知道多维数组的元素总数:
const
  arr: array[0..1, 0..2, 0..3] of Integer = 
    (((1,1,1,1), (2,2,2,2), (3,3,3,3)), ((4,4,4,4), (5,5,5,5), (6,6,6,6)));
begin
  ShowMessage(IntToStr(SizeOf(arr) div SizeOf(Integer))); {24}
end;

posted on 2008-04-23 22:26  万一  阅读(...)  评论(...编辑  收藏