函數返回類型的合一可能涉及Void類型（第2.1.5），需要一個清楚的定義什么和Void統一。Void描述一個缺乏的類型，它不能分配為任何其他類型，即使是Dynamic 。這就是說如果一個函數顯式聲明返回 Dynamic，它不能返回 Void。 > Uni?cation of function return types may involve the Void-type (2.1.5) and requires a clear de?nition of what uni?es with Void. With Void describing the absence of a type,it is not assignable to any other type, not even Dynamic. This means that if a function is explicitly declared as returning Dynamic, it cannot return Void. 相反同樣適用：如果一個函數聲明一個Void類型返回，它不能返回Dynamic或者其他任何類型。然而，合一的方向在分配函數的類型時是允許的。 > The opposite applies as well: If a function declares a return type of Void, it cannot return Dynamic or any other type. However, this direction of uni?cation is allowed when assigning function types: ~~~ var func:Void->Void = function() return "foo"; ~~~ 右手函數顯然是Void->String 類型，但我們可以分配它到Void->Void類型的變量 func 。這是因為編譯器可以安全的假定返回類型是不相干的，因為它不能被分配到任何 非Void 類型。 > The right-hand function clearly is of type Void->String,yet we can assign it to the variable func of type Void->Void. This is because the compiler can safely assume that the return type is irrelevant, given that it could not be assigned to any non-Void type.