为什么我（有时候）必须引用我所引用的程序集引用的程序？ [英] Why do I (sometimes) have to reference assemblies referenced by the assembly I reference?

问题描述

我有一个定义了一些重载接口的组件答：

I have an assembly A that defines an interface with some overloads:

public interface ITransform
{
    Point InverseTransform(Point point);
    Rect InverseTransform(Rect value);
    System.Drawing.Point InverseTransform(System.Drawing.Point point);
}

...并引用A（二进制，而不是项目），并调用重载之一的程序集B：

...and an assembly B that references A (the binary, not the project) and calls one of the overloads:

var transform =
    (other.Source.TransformToDisplay != null &&
    other.Source.TransformToDisplay.Valid) ?
    other.Source.TransformToDisplay : null;
if (transform != null)
{
    e.Location = transform.InverseTransform(e.Location);
}

要为precise，它调用 InverseTransform 方法的 System.Windows.Point 过载，因为那是财产位置在电子的类型。

To be precise, it calls the System.Windows.Point overload of the InverseTransform method, because that is the type of the property Location in e.

但是，当我在IDE中建立B I得到：

But when I build B in the IDE I get:

错误CS0012：类型'System.Drawing.Point'在未引用的程序集定义。您必须添加一个引用到组装'System.Drawing中，版本= 4.0.0.0，文化=中性公钥= b03f5f7f11d50a3a'。

error CS0012: The type 'System.Drawing.Point' is defined in an assembly that is not referenced. You must add a reference to assembly 'System.Drawing, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a'.

尽管这甚至不是我呼吁过载。当我注释掉，其中重载方法 InverseTransform 被称为线，它建立精细即使我仍然实例类型的对象 ITransform 。

even though that's not even the overload I am calling. When I comment out the line where the overloaded method InverseTransform is called, it builds fine even though I'm still instantiating an object of type ITransform.

为什么呢？而且是有办法解决这个问题，而无需添加引用 System.Drawing中无处不在？

Why? And is there a way to fix this without having to add a reference to System.Drawing everywhere?

推荐答案

编译器需要知道什么是 System.Drawing.Point 是为了证明这不是正确的过载（例如，如果它有一个隐式转换）。

The compiler needs to know what a System.Drawing.Point is in order to prove that it's not the correct overload (eg, if it has an implicit conversion).

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