有没有简洁的方法来更新记录的TDictionary中的所有值？

Question

我想做这样的事情,但它不会编译因为无法分配对.

var
  MyDictionary: TDictionary<TGuid, TCustomRecord>;
  Pair: TPair<TGuid, TCustomRecord>;
begin
  // ... create and populate my dictionary ...

  foreach Pair in MyDictionary do
  begin
    PairRef.Value.MyField := PairRef.Value.MyField + 1;
  end;
end

为了清楚起见,我知道如何通过更多代码实现这一目标,我正在寻找简洁易读的东西.

Answer 1

这是一个简单的程序,它显示了使用记录和对象的不同处理方式TDictionary.

program Project1;

{$APPTYPE CONSOLE}
{$R *.res}

uses
  System.SysUtils, System.Generics.Collections;

type
  TMyRecord = record
    Field : Integer;
  end;

  TMyObject = class
    Field : Integer;
  end;

procedure UseObjectDict;
var
  LDict :  TDictionary<TGUID, TMyObject>;
  LValue : TMyObject;
begin
  write( 'TMyObject: ' );

  LDict := TObjectDictionary<TGUID, TMyObject>.Create( [doOwnsValues] );
  try

    // populate
    while LDict.Count < 10 do
    begin
      LDict.Add( TGuid.NewGuid, TMyObject.Create );
    end;

    // update
    for LValue in LDict.Values do
      begin
        LValue.Field := LValue.Field + 1;
      end;

    // output
    for LValue in LDict.Values do
      begin
        write( LValue.Field, ', ' );
      end;
    Writeln;

  finally
    LDict.Free;
  end;
end;

procedure UseRecordDict;
var
  LDict :  TDictionary<TGUID, TMyRecord>;
  LKey :   TGUID;
  LValue : TMyRecord;
begin
  write( 'TMyRecord: ' );
  LDict := TDictionary<TGUID, TMyRecord>.Create;
  try

    // populate
    while LDict.Count < 10 do
      begin
        LValue.Field := 0;
        LDict.Add( TGuid.NewGuid, LValue );
      end;

    // update
    for LKey in LDict.Keys do
      begin
        LValue.Field := LDict[LKey].Field + 1;
        LDict.AddOrSetValue( LKey, LValue );
      end;

    // output
    for LValue in LDict.Values do
      begin
        write( LValue.Field, ', ' );
      end;
    Writeln;

  finally
    LDict.Free;
  end;
end;

begin
  ReportMemoryLeaksOnShutdown := True;
  try

    UseObjectDict;
    UseRecordDict;

  except
    on E : Exception do
      Writeln( E.ClassName, ': ', E.Message );
  end;

  ReadLn;

end.