我正在尝试根据函数类型参数实例化一个类。
虽然documentation说这是可能的,我不能让它工作。
考虑以下代码:
// Dialog base class
// Every dialog in my application will derive from this
class Dialog
{
public function new()
{
// do some stuff here
}
}
// One of the possible dialogs in the application
// Extends Dialog
class TestDialog extends Dialog
{
public function new()
{
super();
// do some more stuff
}
}
// A simple class that tries to instantiate a specialized dialog, like TestDialog
class SomeAppClass
{
public function new()
{
var instance = create(TestDialog);
}
@:generic
function create<T:Dialog>(type:Class<T>):T
{
return new T();
}
}
这不适用于以下错误:
create.T does not have a constructor
显然,我做错了什么,但是什么?
最佳答案
SpecialDialog
可能具有与 Dialog
不同的构造函数.
所以你必须约束它 然后也约束到 Dialog
.
Code @ Try Haxe
package;
typedef Constructible = {
public function new():Void;
}
// Dialog base class
// Every dialog in my application will derive from this
class Dialog
{
public function new()
{
trace("dialog");
}
}
class SuperDialog extends Dialog
{
public function new()
{
super();
trace("super dialog");
}
}
// A simple class that tries to instantiate a specialized dialog, like TestDialog
class SomeAppClass
{
public function new()
{
var dialog = create(Dialog);
var superDialog = create(SuperDialog);
}
@:generic
public static function create<T:(Constructible,Dialog)>(type:Class<T>):T
{
return new T();
}
}
class Test {
static public function main() {
new SomeAppClass();
}
}
关于constructor - Haxe中泛型类型参数的构建,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30609852/