求解这个 eval 函数返回值为什么是个 lambda？怎么才能拿到值？

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 class Min { override fun eval(variables: Map<String, Any>): Any = { val values = parameters.map { it.eval(variables) } min(values[0] as Double, values[1] as Double) } }

这个函数啊在 IDEA 的 evaluate 中执行返回了 Min$eval$1 类的实例 ()-> Kotlin.Double

 override fun eval(variables: Map<String, Any>): Any { val values = parameters.map { it.eval(variables) } return min(values[0] as Double, values[1] as Double) }

我把它改成了普通的 return 函数就没这个问题了可以返回具体的值

求解这种直接等于号赋值的操作和普通函数使用 return 的操作有什么区别

4 replies 2021-05-17 15:43:21 +08:00

AoEiuV020

May 17, 2021

前者=相当于返回=后面的东西，{}花括号包起来那就是个 lambda 了，
override fun eval(variables: Map<String, Any>): Any {
return {
val values = parameters.map { it.eval(variables) }
min(values[0] as Double, values[1] as Double)
}
}

yujincheng08

May 17, 2021

fun x() = blablabla 就是 fun x() { return blablabla } 的语法糖

你 fun x() = { xxx } 就是 fun x() { return () -> { xxx } }

AoEiuV020

May 17, 2021

如果能改成链式调用的话可以考虑用=，
override fun eval(variables: Map<String, Any>): Any = parameters.subList(0, 2).minBy { it.eval(variables) as Double }

AoEiuV020

May 17, 2021

不对，不该 minBy，
override fun eval(variables: Map<String, Any>): Any = parameters.subList(0, 2).map { it.eval(variables) as Double }.min()