在我继续学习scala的过程中,我正在阅读Odersky的《Scala by example》一书,并在第一类函数章节中工作。匿名函数部分避免了递归匿名函数的情况。我有一个看起来可行的解决方案,但我很好奇是否有更好的答案。从pdf中可以看到,这是展示高阶函数的代码。
从PDF中: 展示匿名函数的代码
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
从PDF中: 展示匿名函数的代码
def sum(f: Int => Int, a: Int, b: Int): Int =
if (a > b) 0 else f(a) + sum(f, a + 1, b)
def 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
我的代码:
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
echo "2^2+3^2" | bc -l
-->13
。 - sarnold2^a + 2^a+1 + .... 2^b-1 + 2^b
2^2+2^3 = 4+8 = 12
- James T Kirk