数组 – 如何在Delphi中合并2个字符串数组

我有2个或更多的动态字符串数组填充一些巨大的数据,我想将这个2数组合并到一个数组,我知道我可以用一个for循环这样做：

var
  Arr1,Arr2,MergedArr: Array of string;
  I: Integer;
begin
  // Arr1:= 5000000 records
  // Arr2:= 5000000 records
 
  // Fill MergedArr by Arr1
  MergedArr:= Arr1;
 
  // Set length of MergedArr to length of ( Arra1 + Arr2 )+ 2
  SetLength(MergedArr,High(Arr1)+ High(Arr2)+2);
 
  // Add Arr2 to MergedArr
  for I := Low(Arr2)+1 to High(Arr2)+1 do
    MergedArr[High(Arr1)+ i]:= Arr2[i-1];
end;

但是在巨大的数据上是缓慢的,是否有更快的方式像复制数组内存数据？

解决方法

您可以使用内置的Move功能将内存块移动到另一个位置.参数是源和目标内存块和要移动的数据的大小.

由于您正在复制字符串,源数组在合并后必须通过将其填充为零来销毁.否则,对于字符串的引用将会在程序中造成破坏和破坏.

var
  Arr1,MergedArr: Array of string;
  I: Integer;
begin
  SetLength(Arr1,5000000);
  for I := Low(Arr1) to High(Arr1) do
    Arr1[I] := IntToStr(I);
 
  SetLength(Arr2,5000000);
  for I := Low(Arr2) to High(Arr2) do
    Arr2[I] := IntToStr(I);
 
  // Set length of MergedArr to length of ( Arra1 + Arr2 )+ 2
  SetLength(MergedArr,High(Arr1)+ High(Arr2)+2);
 
  // Add Arr1 to MergedArr
  Move(Arr1[Low(Arr1)],MergedArr[Low(MergedArr)],Length(Arr1)*SizeOf(Arr1[0]));
 
  // Add Arr2 to MergedArr
  Move(Arr2[Low(Arr2)],MergedArr[High(Arr1)+1],Length(Arr2)*SizeOf(Arr2[0]));
 
  // Cleanup Arr1 and Arr2 without touching string refcount.
  FillChar(Arr1[Low(Arr1)],Length(Arr1)*SizeOf(Arr1[0]),0);
  FillChar(Arr2[Low(Arr2)],Length(Arr2)*SizeOf(Arr2[0]),0);
 
  // Test
  for I := Low(Arr1) to High(Arr1) do begin
    Assert(MergedArr[I] = IntToStr(I));
    Assert(MergedArr[I] = MergedArr[Length(Arr1) + I]);
  end;
 
  // Clear the array to see if something is wrong with refcounts
  for I := Low(MergedArr) to High(MergedArr) do
    MergedArr[I] := '';
end;

数组 – 如何在Delphi中合并2个字符串数组

解决方法

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