coderstory
V2EX  ›  Kotlin

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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  •   coderstory · May 17, 2021 · 2830 views
    This topic created in 1846 days ago, the information mentioned may be changed or developed.
       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
        1
    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
        2
    yujincheng08  
       May 17, 2021   ❤️ 1
    fun x() = blablabla 就是 fun x() { return blablabla } 的语法糖

    你 fun x() = { xxx } 就是 fun x() { return () -> { xxx } }
    AoEiuV020
        3
    AoEiuV020  
       May 17, 2021   ❤️ 1
    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
        4
    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()
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