Question

我正在尝试使用SICP学习Scala，但我很难对函数的类型定义感到困惑并陷入SICP。这里构建一个通用表达式来查找数字的平方根（通过定点搜索或牛顿方法）而不是：

def sqrt_damp(x: Double) =
  fixed_point(average_damp(y => x / y))(1)

def sqrt_newton(x: Double) =
  fixed_point(newton_method(y => square(y) - x))(1)

基于功能：

def square(x: Double) = x * x           
def average(x: Double, y: Double) = (x + y) / 2                                             
def abs(x: Double) = if (x < 0) -x else x 

val tolerance = 0.00001                   
def fixed_point(f: Double => Double)(first_guess: Double) = {
  def close_enough(v1: Double, v2: Double): Boolean = abs(v1 - v2) < tolerance
  def attempt(guess: Double): Double = {
    val next = f(guess)
    if (close_enough(guess, next)) next else attempt(next)
  }
  attempt(first_guess)
}

def average_damp(f: Double => Double): Double => Double =
  x => average(x, f(x))

val dx = 0.00001                                
def deriv(g: Double => Double): Double => Double =
  x => (g(x + dx) - g(x)) / dx

def newton_transform(g: Double => Double): Double => Double =
  x => x - g(x) / deriv(g)(x)               

def newton_method(g: Double => Double)(guess: Double): Double =
    fixed_point(newton_transform(g))(guess)

方形函数可以通过以下形式推广：

(define (fixed-point-of-transform g transform guess)
  (fixed-point (transform g) guess))

我试图在Scala中表达如下：

 def fixed_point_of_transform(g: Double => Double, transform: Double => Double)(guess: Double): Double =
   fixed_point(transform(g))(guess)

但是上面没有编译并生成错误

type mismatch; found : Double => Double required: Double

编辑，以下作品：

def fixed_point_of_transform(g: Double => Double, transform: (Double => Double) => (Double => Double))(guess: Double): Double =
  fixed_point(transform(g))(guess)

所以现在以前的功能可以定义为：

def sqrt_damp(x: Double) =
    fixed_point_of_transform(y => x / y, average_damp)(1)

def sqrt_newton(x: Double) =
    fixed_point_of_transform(y => square(y) - x, newton_method)(1)

Answer 1

transform需要Double并返回Double。您无法将其应用于g，因为g是一个函数Double => Double。您可以将其应用于g(x)，x: Double。我想这就是你想要的：fixed_point((x: Double) => transform(g(x)))(guess)

接受函数的函数中的参数类型

1 个答案: