为什么TypeScript允许方法参数中的隐式向下转换？ [重复]

Question

此问题已经在这里有了答案：

why can in TypeScript a possible number value in an interface be converted to a not possible number value in a class implementation?（1个答案）

关闭57秒前。

这里是示例：

interface Vehicle{
    mass:number
}
interface InspectorClass{
    inspect(v:Vehicle):void
}


class Car implements Vehicle{
    mass = 2000
    wheels = 4
}
class Boat implements Vehicle{
    mass = 3000
    sails = 2
}


// with methods it silently fails:

class BoatInspector implements InspectorClass{
    inspect(v:Boat){ // interface contract silently violated!!!
        console.log(v.mass)
        console.log(v.sails)
    }
}

function runInspection(inspector:InspectorClass, vehicle:Vehicle){
    inspector.inspect(vehicle)
}

let myCar = new Car()
let myBoatInspector = new BoatInspector()
runInspection(myBoatInspector, myCar)


// with functions it checks properly:

type InspectorFunction = (v:Vehicle) => void

const inspectCar:InspectorFunction = function(v:Car){ // TypeScript complains as it should
    console.log(v.mass)
    console.log(v.wheels)
}

TypeScript Playground

接口的合同规定，InspectorClass实例中的检查方法必须能够检查任何类型的Vehicle。为什么TypeScript可以让我实现一个实际上只接受Boats而不会抱怨的类？这是错误吗？还是出于某种原因而设计？还是可以通过一些标志启用它？

Answer 1

链接的重复项说明了这种情况。对于上面的代码，解决方法可能是使用--strictFunctionTypes并将方法签名写为函数值属性签名：

interface InspectorClass{
    inspect: (v:Vehicle) => void
}

Playground link