问题

使用以下方法可以创建未指定“深度”的深度分支结构：

TGItem = record
  NameShort: String;
  NameLong: String;
  Formula: String;
  Components: TArray<TGItem>;
  procedure Init;
end;

var 
 A, B: TGItem;

 B := A; 

在A和B的.Components都指向相同的数组，但我想复制数组。如何做到这一点？

Version 10.4

Edit/Current progress:

class operator Assign(var Dest: TGItem; const [ref] Src: TGItem);

以及

class operator TGItem.Assign(var Dest: TGItem; const [ref] Src: TGItem);
begin
  // 在这里放什么？
end;

TGItem = record
  NameShort:            String;
  NameLong:             String;
  Formula:              String;
  Components:           TArray<TGItem>;
  procedure Init;
end;

var 
 A,B: TGItem;

 B := A;  

class operator Assign(var Dest: TGItem; const [ref] Src: TGItem);

class operator TGItem.Assign(var Dest: TGItem; const [ref] Src:TGItem);
begin
  // What to put here?
end;

答案1

得分: 1

function TGItem.Clone: TGItem;
begin
  Result.Init;    
  Result.NameShort        := Self.NameShort;
  Result.NameLong         := Self.NameLong;
  Result.Formula          := Self.Formula;
  Setlength(Result.Components, Length(Self.Components));
    
  var I: Integer;
  for I := 0 to Length(Result.Components)-1 do
  begin
    Result.Components[I] := Self.Components[I].Clone;
  end;    
end;

似乎有效，而不是使用 B := A，我使用 B := A.Clone，它执行递归复制并创建一个新的克隆的分支结构（而不是指向旧结构）。

<details>
<summary>英文:</summary>

function TGItem.Clone: TGItem;
begin
Result.Init;
Result.NameShort := Self.NameShort;
Result.NameLong := Self.NameLong;
Result.Formula := Self.Formula;
Setlength(Result.Components, Length(Self.Components));

var I: Integer;
for I := 0 to Length(Result.Components)-1 do
begin
Result.Components[I] := Self.Components[I].Clone;
end;
end;

Seems to work, instead of `B := A` I use `B := A.Clone` and it does a recursive copy and creates a new cloned branching structure (instead of pointing to the old structure).

</details>