Question

鉴于我们在作用域中有两个函数，它们的名称相同但参数列表不同。如何区分彼此，例如在尝试访问tupled函数的情况下

示例：

scala> :paste
// Entering paste mode (ctrl-D to finish)


def f(s: String, l: Long): String = ???
def f(i: Int, l: Long): String = ???

val t:((String, Long)) => String = f.tupled

// Exiting paste mode, now interpreting.

<pastie>:15: error: ambiguous reference to overloaded definition,
both method f of type (i: Int, l: Long)String
and  method f of type (s: String, l: Long)String
match expected type ?
val t:((String, Long)) => String = f.tupled

将问题简化为函数文字可得出：

scala> :paste
// Entering paste mode (ctrl-D to finish)

def f(i: Int, l: Long): String = ???
def f(s: String, l: Long): String = ???

val g = f _

// Exiting paste mode, now interpreting.

<pastie>:15: error: ambiguous reference to overloaded definition,
both method f of type (s: String, l: Long)String
and  method f of type (i: Int, l: Long)String
match expected type ?
val g = f _
        ^

但是函数文字示例中的显式类型注释设法解决了这个问题，至于上述apply上下文中的tupled方法不起作用

scala>

scala> :paste
// Entering paste mode (ctrl-D to finish)

def f(s: String, l: Long): String = ???
def f(i: Int, l: Long): String = ???

val g: (String, Long) => String  = f _

// Exiting paste mode, now interpreting.

f: (s: String, l: Long)String <and> (i: Int, l: Long)String
f: (s: String, l: Long)String <and> (i: Int, l: Long)String
g: (String, Long) => String = $$Lambda$1458/794413935@60cbba57

Answer 1

这是您实现这一目标的方法

val t1 = (f(_: String, _: Long)).tupled
val t2 = (f(_: Int, _: Long)).tupled

输出

t1: ((String, Long)) => String = scala.Function2$$Lambda$1188/922196321@3994b698
t2: ((Int, Long)) => String = scala.Function2$$Lambda$1188/922196321@4249db51

Answer 2

我发现了针对tupled问题的以下破解方法：

scala> :paste
// Entering paste mode (ctrl-D to finish)

def f(s: String, l: Long): String = ???
def f(i: Int, l: Long): String = ???

val g: (String, Long) => String  = f _
val t = g.tupled

// Exiting paste mode, now interpreting.

f: (s: String, l: Long)String <and> (i: Int, l: Long)String
f: (s: String, l: Long)String <and> (i: Int, l: Long)String
g: (String, Long) => String = $$Lambda$1476/1981627424@172f2717
t: ((String, Long)) => String = scala.Function2$$Lambda$259/530042637@bda4cbe

如何引用作用域中两个函数具有相同名称的特定函数

2 个答案: