假设我们有一个类和一个重载函数:
public class Main {
static final class A {
}
public static String g(ToIntFunction<? extends A> f) {
return null;
}
public static String g(ToDoubleFunction<? extends A> f) {
return null;
}
}
我想用一个方法引用来调用 g,方法引用类型为 A -> int 的函数:
public class Main {
static final class A {
}
public static String g(ToIntFunction<? extends A> f) {
return null;
}
public static String g(ToDoubleFunction<? extends A> f) {
return null;
}
private static int toInt(A x) {
return 2;
}
public static void main(String[] args) {
ToIntFunction<? extends A> f1 = Main::toInt;
ToDoubleFunction<? extends A> f2 = Main::toInt;
g(Main::toInt);
}
}
这适用于 javac,但不适用于 eclipse ecj。我向 ecj 提交了错误报告,但我不确定这是 ecj 错误还是 javac 错误,并尝试按照重载解析算法来解决这个问题。我的感觉是代码应该被接受,因为 ToIntFunction
在直觉上是合理的更适合 toInt
比ToDoubleFunction
.然而,我对 JLS 的解读是,它应该被拒绝,因为没有理由更具体。
我不得不承认我对 JLS 规范有点迷茫,希望得到一些帮助。我首先想计算 Main::double2int
的类型,所以我看了看15.13.2. Type of a Method Reference .它没有定义类型,但它定义了类型在不同上下文中何时兼容:
A method reference expression is compatible in an assignment context, invocation context, or casting context with a target type T if T is a functional interface type (§9.8) and the expression is congruent with the function type of the ground target type derived from T.
地面类型是ToIntFunction<A>
和 ToDoubleFunction<A>
. toInt
返回一个赋值与 double 兼容的 int,因此我得出结论,该方法引用与 ToIntFunction<? extends A>
兼容和 ToDoubleFunction<? extends A>
在调用上下文中。这可以通过将方法引用分配给 ToIntFunction<? extends A>
来验证。和 ToDoubleFunction<? extends A>
这在主要功能中被接受。
然后我查看重载解析并找到15.12.2.5. Choosing the Most Specific Method其中有一个方法引用的特殊情况来决定 ToIntFunction
的两个重载中的哪一个或 ToDoubleFunction
对于参数 Main::toInt
更具体编译时声明 A -> int
.
A functional interface type S is more specific than a functional interface type T for an expression e if T is not a subtype of S and one of the following is true (where U1 ... Uk and R1 are the parameter types and return type of the function type of the capture of S, and V1 ... Vk and R2 are the parameter types and return type of the function type of T):
...
If e is an exact method reference expression (§15.13.1), then i) for all i (1 ≤ i ≤ k), Ui is the same as Vi, and ii) one of the following is true:
R2 is void.
R1 <: R2.
R1 is a primitive type, R2 is a reference type, and the compile-time declaration for the method reference has a return type which is a primitive type.
R1 is a reference type, R2 is a primitive type, and the compile-time declaration for the method reference has a return type which is a reference type.
第一个条件显然不匹配,因为R1和R2不void。
两个接口(interface)ToIntFunction
和 ToDoubleFunction
不同之处仅在于它们的返回类型是原始类型 double
和 int
.对于原始类型,“R1 <: R2”子句根据类型的大小在 4.10.1 中定义。 double
之间没有关系和 int
,所以这种情况没有定义哪种类型更具体。
后两点也不符合,因为这两个函数式接口(interface)都没有引用类型的返回值。
当两个功能接口(interface)返回原语并且代码应该被拒绝为歧义时,似乎没有规则。但是,javac 接受代码,我希望它这样做。所以我想知道这是否是 JLS 中的一个缺失点。
最佳答案
For primitive types, the clause "R1 <: R2" is defined in 4.10.1 according to the size of the types. There is no relation between double and int, so this case does not define which type is more specific.
事实并非如此; double
实际上是 int
的父类(super class)型。
The supertypes of a type are obtained by reflexive and transitive closure over the direct supertype relation, written
S >₁ T
, which is defined by rules given later in this section. We writeS :> T
to indicate that the supertype relation holds betweenS
andT
.
The following rules define the direct supertype relation among the primitive types:
double >₁ float
float >₁ long
long >₁ int
父类(super class)型 关系是直接父类(super class)型 关系的自反和传递闭包,这意味着从 (double >₁ float) ∧ (float >₁ long) ∧ (long >₁ int)
跟随 double :> int
。
关于java - 使用方法引用和原始类型的函数接口(interface)特化的重载解析,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45878208/