接受函数的函数中参数的类型

Question

我正在尝试使用SICP学习Scala，但是我很难用函数的类型定义并卡在SICP。这里广义表达是构建的，其中代替寻找一个数的平方根（通过定点搜索或牛顿法）：

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) 

基于该功能：

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) 

的平方函数

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

其中我试图表达在斯卡拉如下：可以在形式广义

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

然而，上述不编译并生成错误

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

编辑，以下工作：

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

所以现在前面的功能可以被定义为：

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。您不能将它应用于g，因为g是一个函数Double => Double。您可以将其应用于g(x)，其中x: Double。我认为这是你想要的：fixed_point((x: Double) => transform(g(x)))(guess)