delphi 判斷一個數組的長度用 Length 還是 SizeOf ?

判斷一個數組的長度用 Length 還是 SizeOf ?
最近發現一些代碼, 甚至有一些專家代碼, 在遍歷數組時所用的數組長度竟然是 SizeOf(arr); 這不合適!

如果是一維數組、且元素大小是一個字節, 這樣用看不出錯誤, 譬如:  
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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)]);
{顯示結果: 10,10,10,10}
end;
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但如果數組元素多於一個字節、或是多維數組的情況下, 就不行了, 舉例:
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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;
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我們倒是可以利用這個原理, 迅速知道多維數組的元素總數:
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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;