如何编写递归匿名函数？
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在继续学习scala的过程中，我正在研究Odersky的“ Scala by example”，以及有关一流函数的章节中，有关匿名函数的部分避免了递归匿名函数的情况。我有一个可行的解决方案。我很好奇是否有更好的答案。
从pdf：
展示高阶功能的代码
def sum(f: Int => Int, a: Int, b: Int): Int =
if (a > b) 0 else f(a) + sum(f, a + 1, b)
def id(x: Int): Int = x
def square(x: Int): Int = x * x
def powerOfTwo(x: Int): Int = if (x == 0) 1 else 2 * powerOfTwo(x-1)
def sumInts(a: Int, b: Int): Int = sum(id, a, b)
def sumSquares(a: Int, b: Int): Int = sum(square, a, b)
def sumPowersOfTwo(a: Int, b: Int): Int = sum(powerOfTwo, a, b)
scala> sumPowersOfTwo(2,3)
res0: Int = 12
def sum(f: Int => Int, a: Int, b: Int): Int =
if (a > b) 0 else f(a) + sum(f, a + 1, b)
def sumInts(a: Int, b: Int): Int = sum((x: Int) => x, a, b)
def sumSquares(a: Int, b: Int): Int = sum((x: Int) => x * x, a, b)
// no sumPowersOfTwo
def sumPowersOfTwo(a: Int, b: Int): Int = sum((x: Int) => {
def f(y:Int):Int = if (y==0) 1 else 2 * f(y-1); f(x) }, a, b)
scala> sumPowersOfTwo(2,3)
res0: Int = 12
没有找到相关结果
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2 个回复
薄响
(new Function1[Int,Unit] { def apply(x: Int) { println(\"\" + x) if (x > 1) this(x - 1) } })(10)val printingCounter: (Int) => Unit = (x: Int) => { println(\"\" + x) if (x > 1) printingCounter(x - 1) } printingCounter(10)淘圃跺枯替
case class Rec[I, O](fn : (I => O, I) => O) extends (I => O) { def apply(v : I) = fn(this, v) }例如，相同的解决方案可用于实现备忘录。