在Delphi中将const数组复制到动态数组

时间:2009-06-11 17:07:46

标签: delphi arrays delphi-2009 const

我有一个固定的常数数组

constAry1: array [1..10] of byte = (1,2,3,4,5,6,7,8,9,10);

和动态数组

dynAry1: array of byte;

将值从 constAry1 复制到 dynAry1 的最简单方法是什么?

如果你有一个数组的const数组(多维),它会改变吗?

constArys: array [1..10] of array [1..10] of byte = . . . . .

3 个答案:

答案 0 :(得分:26)

这会将 constAry1 复制到 dynAry

SetLength(dynAry, Length(constAry1));
Move(constAry1[Low(constAry1)], dynAry[Low(dynAry)], SizeOf(constAry1));

答案 1 :(得分:12)

function CopyByteArray(const C: array of Byte): TByteDynArray;
begin
  SetLength(Result, Length(C));
  Move(C[Low(C)], Result[0], Length(C));
end;

procedure TFormMain.Button1Click(Sender: TObject);
const
  C: array[1..10] of Byte = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10);
var
  D: TByteDynArray;
  I: Integer;
begin
  D := CopyByteArray(C);
  for I := Low(D) to High(D) do
    OutputDebugString(PChar(Format('%d: %d', [I, D[I]])));
end;

procedure TFormMain.Button2Click(Sender: TObject);
const
  C: array[1..10, 1..10] of Byte = (
    (1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
    (1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
    (1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
    (1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
    (1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
    (1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
    (1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
    (1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
    (1, 2, 3, 4, 5, 6, 7, 8, 9, 10),
    (1, 2, 3, 4, 5, 6, 7, 8, 9, 10));

var
  D: array of TByteDynArray;
  I, J: Integer;
begin
  SetLength(D, Length(C));
  for I := 0 to Length(D) - 1 do
    D[I] := CopyByteArray(C[Low(C) + I]);

  for I := Low(D) to High(D) do
    for J := Low(D[I]) to High(D[I]) do
      OutputDebugString(PChar(Format('%d[%d]: %d', [I, J, D[I][J]])));
end;

答案 2 :(得分:2)

从Delphi XE7开始,允许对数组使用类似字符串的操作。然后,您可以直接声明动态数组的常量。例如:

const
  KEY: TBytes = [$97, $45, $3b, $3e, $c8, $14, $c9, $e1];