我可以在Objective-C接口中声明'protocol'类型的变量吗？

My idea is very similar to declare a variable of an interface type in java.

我的想法非常类似于在java中声明一个接口类型的变量。

So for example,

例如,

header file 1:

头文件1:

@protocol Calculator

@end

I then define an @interface CalculatorImpl which implements the above Calculator protocol.

然后我定义了一个@interface CalculatorImpl,它实现了上面的Calculator协议。

In header file 2:

在头文件2中:

@interface SomeViewController : UIViewController {


}

@property (weak, nonatomic) IBOutlet UITextField *txtResult;
@property (weak, nonatomic) Calculator* calculator;

@end

However, the xcode will flag an error at the calculator line

但是,xcode会在计算器行标记错误

property with 'weak' attribute must be of object type

Is this usage of protocol disallowed by objective-c?

Objective-c是否禁止使用协议?

2 个解决方案

#1

A @protocol isn't a type so you can't use it for the type of a @property.

@protocol不是一个类型,所以你不能将它用作@property的类型。

What you probably want to do instead is this:

您可能想要做的是:

@property (weak, nonatomic) id <Calculator> calculator;

This declares a property with no restriction on its type, except that it conforms to the Calculator protocol.

这声明了一个属性,对其类型没有限制,只不过它符合Calculator协议。

#2

You should use

你应该用

@property (weak, nonatomic) id <Calculator> calculator;

In Objective-C you cannot instantiate a protocol, you can only be conform to it. Thus, instead of having an object of type Calculator, you should have a generic object that is conform to Calculator protocol.

在Objective-C中,您无法实例化协议,您只能遵守它。因此,您应该拥有一个符合Calculator协议的通用对象,而不是具有Calculator类型的对象。

Otherwise you can use

否则你可以使用

@property (weak, nonatomic) CalculatorImpl* calculator;

since CalculatorImpl is an interface, not a protocol.

因为CalculatorImpl是一个接口,而不是一个协议。

#1