6

当我使用 TObjectDictionary（其中 TKey 是对象）时，我的应用程序无法正常工作。我有两个单元，那包含两个类。第一单元：

unit RubTerm;

interface

type
  TRubTerm = Class(TObject)
  private
    FRubricName: String;
    FTermName: String;
  public
    property RubricName: String read FRubricName;
    property TermName: String read FTermName;
    constructor Create(ARubricName, ATermName: String);
  end;

implementation

constructor TRubTerm.Create(ARubricName, ATermName: String);
begin
  Self.FRubricName := ARubricName;
  Self.FTermName := ATermName;
end;

end;

第二单元：

unit ClassificationMatrix;

interface

uses
  System.Generics.Collections, System.Generics.Defaults, System.SysUtils, RubTerm;

type
TClassificationMatrix = class(TObject)
  private
    FTable: TObjectDictionary<TRubTerm, Integer>;
  public
    constructor Create;
    procedure TClassificationMatrix.AddCount(ADocsCount: Integer; ARubName, ATermName: String);
    function TClassificationMatrix.GetCount(ARubName, ATermName: String): Integer;
  end;

implementation

constructor TClassificationMatrix.Create;
begin
  FTable := TObjectDictionary<TRubTerm, Integer>.Create;
end;

procedure TClassificationMatrix.AddCount(ADocsCount: Integer; ARubName, ATermName: String);
var
  ARubTerm: TRubTerm;
begin
  ARubTerm := TRubTerm.Create(ARubName, ATermName);
  FTable.Add(ARubTerm, ADocsCount);
end;

function TClassificationMatrix.GetCount(ARubName, ATermName: String): Integer;
var
  ARubTerm: TRubTerm;
begin
  ARubTerm := TRubTerm.Create(ARubName, ATermName);
  FTable.TryGetValue(ARubTerm, Result);
end;

end;

但是这段代码工作异常：

procedure TestTClassificationMatrix.TestGetCount;
var
  DocsCountTest: Integer;
begin
  FClassificationMatrix.AddCount(10, 'R', 'T');
  DocsCountTest := FClassificationMatrix.GetCount('R', 'T');
end;
// DocsCountTest = 0! Why not 10? Where is problem?

谢谢！

4

2 回答 2

8

这里的基本问题是您的类型的默认相等比较器的行为方式与您希望的方式不同。您希望相等表示值相等，但默认比较给出引用相等。

您希望值相等这一事实强烈表明您应该使用值类型而不是引用类型。这是我建议的第一个改变。

type
  TRubTerm = record
    RubricName: string;
    TermName: string;
    class function New(const RubricName, TermName: string): TRubTerm; static;
    class operator Equal(const A, B: TRubTerm): Boolean;
    class operator NotEqual(const A, B: TRubTerm): Boolean;
  end;

class function TRubTerm.New(const RubricName, TermName: string): TRubTerm;
begin
  Result.RubricName := RubricName;
  Result.TermName := TermName;
end;

class operator TRubTerm.Equal(const A, B: TRubTerm): Boolean;
begin
  Result := (A.RubricName=B.RubricName) and (A.TermName=B.TermName);
end;

class operator TRubTerm.NotEqual(const A, B: TRubTerm): Boolean;
begin
  Result := not (A=B);
end;

我添加TRubTerm.New了一个帮助方法，以便于初始化记录的新实例。并且为方便起见，您可能还会发现重载等式和不等式运算符很有用，就像我在上面所做的那样。

切换到值类型后，您还将更改字典以匹配。使用TDictionary<TRubTerm, Integer>而不是TObjectDictionary<TRubTerm, Integer>. 切换到值类型还有助于修复现有代码中的所有内存泄漏。您现有的代码创建对象但从不破坏它们。

这会让你回家，但你仍然需要为你的字典定义一个相等比较器。记录的默认比较器将基于引用相等性，因为字符串尽管表现为值类型，但仍存储为引用。

要制作合适的相等比较器，您需要实现以下比较函数，其中T替换为TRubTerm：

TEqualityComparison<T> = reference to function(const Left, Right: T): Boolean;
THasher<T> = reference to function(const Value: T): Integer;

我会将这些实现为记录的静态类方法。

type
  TRubTerm = record
    RubricName: string;
    TermName: string;
    class function New(const RubricName, TermName: string): TRubTerm; static;
    class function EqualityComparison(const Left, 
      Right: TRubTerm): Boolean; static;
    class function Hasher(const Value: TRubTerm): Integer; static;
    class operator Equal(const A, B: TRubTerm): Boolean;
    class operator NotEqual(const A, B: TRubTerm): Boolean;
  end;

实施EqualityComparison很容易：

class function TRubTerm.EqualityComparison(const Left, Right: TRubTerm): Boolean;
begin
  Result := Left=Right;
end;

但是哈希器需要更多的思考。您需要单独散列每个字段，然后组合散列。以供参考：

代码如下所示：

{$IFOPT Q+}
  {$DEFINE OverflowChecksEnabled}
  {$Q-}
{$ENDIF}
function CombinedHash(const Values: array of Integer): Integer;
var
  Value: Integer;
begin
  Result := 17;
  for Value in Values do begin
    Result := Result*37 + Value;
  end;
end;
{$IFDEF OverflowChecksEnabled}
  {$Q+}
{$ENDIF}

function GetHashCodeString(const Value: string): Integer;
begin
  Result := BobJenkinsHash(PChar(Value)^, SizeOf(Char) * Length(Value), 0);
end;

class function TRubTerm.Hasher(const Value: TRubTerm): Integer;
begin
  Result := CombinedHash([GetHashCodeString(Value.RubricName), 
    GetHashCodeString(Value.TermName)]);
end;

最后，当你实例化你的字典时，你需要提供一个IEqualityComparison<TRubTerm>. 像这样实例化你的字典：

Dict := TDictionary<TRubTerm,Integer>.Create(
  TEqualityComparer<TRubTerm>.Construct(
    TRubTerm.EqualityComparison,
    TRubTerm.Hasher
  )
);
于 2013-08-06T08:31:00.467 回答
3

字典取决于键值。您正在密钥中存储对对象的引用。如果您创建两个设置相同的对象，则它们具有不同的值，因此具有不同的键。

var
  ARubTerm1: TRubTerm;
  ARubTerm2: TRubTerm;
begin
  ARubTerm1 := TRubTerm.Create('1', '1');
  ARubTerm2 := TRubTerm.Create('1', '1');
 //  ARubTerm1 = ARubTerm2 is not possible here as ARubTerm1 points to a different address than ARubTerm2
end;

相反，您可以使用字符串作为基于 RubricName 和 TermName 的 TObjectDictonary 中的第一个类型参数。有了这个，你就会得到相同的值。

还应注意，上述 XE2 中的代码会产生两个内存泄漏。创建的每个对象都必须被释放。因此，这部分代码也正在泄漏内存

function TClassificationMatrix.GetCount(ARubName, ATermName: String): Integer;
var
  ARubTerm: TRubTerm;
begin
  ARubTerm := TRubTerm.Create(ARubName, ATermName);
  FTable.TryGetValue(ARubTerm, Result);
end;

鉴于这一切。如果您想使用对象作为键，您可以使用自定义相等比较器来完成。这是您的示例更改为实现IEqualityComparer<T>，并修复了一些内存泄漏。

unit ClassificationMatrix;

interface

uses
  Generics.Collections, Generics.Defaults, SysUtils, RubTerm;

type
TClassificationMatrix = class(TObject)
  private
    FTable: TObjectDictionary<TRubTerm, Integer>;
  public
    constructor Create;
    procedure AddCount(ADocsCount: Integer; ARubName, ATermName: String);
    function GetCount(ARubName, ATermName: String): Integer;
  end;

implementation

constructor TClassificationMatrix.Create;
var
 Comparer : IEqualityComparer<RubTerm.TRubTerm>;
begin
  Comparer := TRubTermComparer.Create;
  FTable := TObjectDictionary<TRubTerm, Integer>.Create([doOwnsKeys],TRubTermComparer.Create);
end;

procedure TClassificationMatrix.AddCount(ADocsCount: Integer; ARubName, ATermName: String);
var
  ARubTerm: TRubTerm;
begin
  ARubTerm := TRubTerm.Create(ARubName, ATermName);
  FTable.Add(ARubTerm, ADocsCount);
end;

function TClassificationMatrix.GetCount(ARubName, ATermName: String): Integer;
var
  ARubTerm: TRubTerm;
begin
  ARubTerm := TRubTerm.Create(ARubName, ATermName);
  try
   if Not FTable.TryGetValue(ARubTerm, Result) then
      result := 0;
  finally
    ARubTerm.Free;
  end;
end;

end.

和 RubTerm.pas 单元

unit RubTerm;

interface
uses Generics.Defaults;

type
  TRubTerm = Class(TObject)
  private
    FRubricName: String;
    FTermName: String;
  public
    property RubricName: String read FRubricName;
    property TermName: String read FTermName;
    constructor Create(ARubricName, ATermName: String);
    function GetHashCode: Integer; override;
  end;

  TRubTermComparer = class(TInterfacedObject, IEqualityComparer<TRubTerm>)
  public
    function Equals(const Left, Right: TRubTerm): Boolean;
    function GetHashCode(const Value: TRubTerm): Integer;
  end;


implementation

constructor TRubTerm.Create(ARubricName, ATermName: String);
begin
  Self.FRubricName := ARubricName;
  Self.FTermName := ATermName;
end;


{ TRubTermComparer }

function TRubTermComparer.Equals(const Left, Right: TRubTerm): Boolean;
begin
  result := (Left.RubricName = Right.RubricName) and (Left.TermName = Right.TermName);
end;

function TRubTermComparer.GetHashCode(const Value: TRubTerm): Integer;
begin
  result := Value.GetHashCode;
end;

//The Hashing code was taken from David's Answer to make this a complete answer.    
{$IFOPT Q+}
  {$DEFINE OverflowChecksEnabled}
  {$Q-}
{$ENDIF}
function CombinedHash(const Values: array of Integer): Integer;
var
  Value: Integer;
begin
  Result := 17;
  for Value in Values do begin
    Result := Result*37 + Value;
  end;
end;
{$IFDEF OverflowChecksEnabled}
  {$Q+}
{$ENDIF}

function GetHashCodeString(const Value: string): Integer;
begin
  Result := BobJenkinsHash(PChar(Value)^, SizeOf(Char) * Length(Value), 0);
end;

function TRubTerm.GetHashCode: Integer;

begin
  Result := CombinedHash([GetHashCodeString(Value.RubricName), 
    GetHashCodeString(Value.TermName)]);    
end;

end.
于 2013-08-06T00:29:44.693 回答