令我惊讶的是,泛型不受“ is”和“ as”运算的约束。
如何实现以下目标?
function TDisplayConfManager._getDisplay<T>(parentCtl: TWinControl; indicatorName: string): T;
var
i: Integer;
ind: T;
begin
for i := 0 to parentCtl.ControlCount-1 do
if parentCtl.Controls[i] is T then begin // E2015 Operator not applicable to this operand type
ind := parentCtl.Controls[i] as T; // E2015 Operator not applicable to this operand type
if SameText(ind.Caption, indicatorName) then Exit(ind); // E2003 Undeclared identifier: 'Caption'
end;
Result := T.Create(_owner); // E2003 Undeclared identifier: 'Create'
end;
如果我对所有显示器都使用基本显示类,则会出现两个错误,但是我仍然无法使用“ is”来创建对象:
function TDisplayConfManager._getDisplay<T>(parentCtl: TWinControl; indicatorName: string): T;
var
i: Integer;
ind: TBaseDisplayType;
begin
for i := 0 to parentCtl.ControlCount-1 do
if parentCtl.Controls[i] is T then begin // E2015 Operator not applicable to this operand type
ind := parentCtl.Controls[i] as TBaseDisplayType;
if SameText(ind.Caption, indicatorName) then Exit(ind);
end;
Result := T.Create(_owner); // E2003 Undeclared identifier: 'Create'
end;
答案 0 :(得分:1)
原因是T
不必是类类型;它也可以是非类类型,例如整数,字符串,记录或数组。
因此以下内容无法编译:
type
TTest<T> = record
function Test(AObject: TObject): Boolean;
end;
{ TTest<T> }
function TTest<T>.Test(AObject: TObject): Boolean;
begin
Result := AObject is T; // E2015 Operator not applicable to this operand type
end;
但是,如果您知道自己的T
类型始终是类类型,则可以在Delphi中表示出来:
type
TTest<T: class> = record
function Test(AObject: TObject): Boolean;
end;
{ TTest<T> }
function TTest<T>.Test(AObject: TObject): Boolean;
begin
Result := AObject is T;
end;
documentation包含有关通用约束的更多详细信息。
根据评论进行更新:
您还可以在成员级别上设置约束:
type
TTest = record
function Test<T: class>(AObject: TObject): Boolean;
end;
{ TTest }
function TTest.Test<T>(AObject: TObject): Boolean;
begin
Result := AObject is T;
end;