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:
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;