Scala如何以不同的方式推断这两个函数?
这包括:Scala如何以不同的方式推断这两个函数?,scala,type-inference,Scala,Type Inference,这包括: def fibonacci():() => Int = { var first = 1 var second = 2 return () => { val oldFirst = first first = second second = second + oldFirst second } } 这
def fibonacci():() => Int = {
var first = 1
var second = 2
return () => {
val oldFirst = first
first = second
second = second + oldFirst
second
}
}
这并不是:
def fibonacci():() => Int = {
var first = 1
var second = 2
return ():Int => {
val oldFirst = first
first = second
second = second + oldFirst
second
}
}
我显式地试图告诉它我正在返回一个Int,但是我得到了这个错误:声明的非法开始
,它指向第一行=第二行。它们有何不同?我正在使用Scala 2.8.1。return():Int=>{…}
在Scala中不是正确的表达式。如果要显式指定返回类型,则需要将声明放在值之后(该值将是匿名函数):
但是,请注意,没有必要这样做。如果省略return
,则根本不需要进行任何显式类型声明:
def fibonacci() = {
var first = 1
var second = 2
() => {
val oldFirst = first
first = second
second = second + oldFirst
second
}
}
return():Int=>{…}
在Scala中不是正确的表达式。如果要显式指定返回类型,则需要将声明放在值之后(该值将是匿名函数):
但是,请注意,没有必要这样做。如果省略return
,则根本不需要进行任何显式类型声明:
def fibonacci() = {
var first = 1
var second = 2
() => {
val oldFirst = first
first = second
second = second + oldFirst
second
}
}
德比尔斯基是对的。另外两条注释:(1)不需要使用return
关键字。默认情况下,最后一个表达式将成为返回值。(2) 如果您试图单独注释函数体的类型,这是可能的。代码变为:
def fibonacci2(): () => Int = {
var first = 1
var second = 2
() => {
val oldFirst = first
first = second
second = second + oldFirst
second
}: Int
}
德比尔斯基是对的。另外两条注释:(1)不需要使用return
关键字。默认情况下,最后一个表达式将成为返回值。(2) 如果您试图单独注释函数体的类型,这是可能的。代码变为:
def fibonacci2(): () => Int = {
var first = 1
var second = 2
() => {
val oldFirst = first
first = second
second = second + oldFirst
second
}: Int
}