假設我有一個父類和兩個這樣的子類abstract class Vehicle { public function setManufacturer(string $manufacturer) { ... }}class Bicycle extends Vehicle { public function addSaddle() { ... }}class Car extends Vehicle { public function setSpeedLimit(float $limit) { ... }}現在我想通過這樣的接口使用這些對象interface VehicleInterface { public function create(Vehicle $vehicle);}class BicycleService implements VehicleInterface { public function create(Bicycle $bicycle) { $bicycle->setManufacturer('Some company'); // Common for all Vehicle objects $bicycle->addSaddle(); // Common only for Bicycle objects }}看來這是不可能的，因為BicycleService create(...)與接口不兼容。除了刪除類型提示之外，還有什么辦法可以解決這個問題嗎？
4 回答
守候你守候我
TA貢獻1802條經驗 獲得超10個贊
考慮一下：
class NotBicycle implements Vehicle { ... }
interface VehicleInterface {
? ? public function create(Vehicle $vehicle);
}
class BicycleService implements VehicleInterface {
? ? public function create(Bicycle $bicycle) {
? ? ? ? $bicycle->setManufacturer('Some company'); // Common for all Vehicle objects
? ? ? ? $bicycle->addSaddle(); // Common only for Bicycle objects
? ? }
}
function createVehicle(VehicleInterface $service, Vehicle $vehicle) {
? ? $service->create($vehicle);
}
$service = new BicycleService();
$vehicle = new NotBicycle();
createVehicle($service, $vehicle);
即使你以某種方式只能接受Bicycle其中的一個BicycleService，createVehicle($service, $vehicle)仍然會起作用，因為VehicleInterface有一個方法create(Vehicle $vehicle)。因此，為了讓它按照您想要的方式工作，您需要從根本上打破界面的本質。
您唯一真正的選擇是添加運行時類型檢查。就像例如
class BicycleService implements VehicleInterface {
? ? public function create(Vehicle $bicycle) {
? ? ? ? if (!$bicycle instance of Bicycle) {?
? ? ? ? ? ?throw new TypeError('Expected Bicycle but got '.get_class($bicycle));?
? ? ? ? }
? ? ? ? $bicycle->setManufacturer('Some company'); // Common for all Vehicle objects
? ? ? ? $bicycle->addSaddle(); // Common only for Bicycle objects
? ? }
}
您嘗試做的事情稱為協變方法參數類型，在大多數面向對象的編程語言中是不允許的，因為它違反了里氏替換原則
30秒到達戰場
TA貢獻1828條經驗 獲得超6個贊
Bicycle您需要刪除帶有 的類型Vehicle。你可能需要檢查if ($bicycle instanceof Bicycle)
抽象層次太抽象了。
慕尼黑的夜晚無繁華
TA貢獻1864條經驗 獲得超6個贊
您需要將您的Vehicle類傳遞給它，然后用于instanceof檢查接口。在下面的例子中，我更改了命名約定。
interface Vehicle {
public function setSpeedLimit(Float $limit) : Car;
public function setManufacturer(String $manufacturer) : MethodOfTransport;
}
interface NonVehicle {
public function addSaddle() : Bicycle;
public function setManufacturer(String $manufacturer) : MethodOfTransport;
}
abstract class MethodOfTransport
{
public function setManufacturer(String $manufacturer) : MethodOfTransport { return $this; }
}
class Bicycle extends MethodOfTransport implements NonVehicle
{
}
class Car extends MethodOfTransport implements Vehicle
{
public function setSpeedLimit(Float $limit) : Car { return $this; }
}
$b = new Bicycle();
$c = new Car();
如前所述，我消除了命名約定的一些混亂，因為 aBicycle不是 aVehicle而是 a MethodOfTransport。
interface VehicleServiceInterface
{
# Change the return DataType to MethodOfTransport if Vehicle implements multiple classes
public function create(MethodOfTransport $t) : Car;
}
class VehicleService implements VehicleServiceInterface
{
public function create(MethodOfTransport $t) : Car
{
if (!$t instanceof Vehicle)
throw new Exception( 'Cannot create a Vehicle on a NonVehicle instance.' );
$t->setManufacturer( 'Foo' )->setSpeedLimit( 140 );
return $t;
}
}
$c = (new VehicleService())->create( $c ); // Works
$b = (new VehicleService())->create( $b ); // Throws Exception
和Car都是Bicyclea MethodOfTransport，但是，一個是 a Vehicle，另一個是 a NonVehicle。兩者都可以傳遞到方法中，但我們可以使用 來確保它是正確的類型instanceof。
但是，您可以將返回類型設置為方法的返回類型，create因為Car我們Vehicle在本例中僅過濾實例。如果您有多個Vehicle已實現的類，請將其更改為返回MethodOfTransport
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