《Kotlin极简教程》第六章 Kotlin函数式编程（FP）

Kotlin对函数式编程的实现恰到好处。
最新上架！！！
《 Kotlin极简教程》 陈光剑 （机械工业出版社）:
https://mp.weixin.qq.com/s/bzRkGSO6T1O2AELM_UqKUQ

函数指针

/** * "Callable References" or "Feature Literals", i.e. an ability to pass * named functions or properties as values. Users often ask * "I have a foo() function, how do I pass it as an argument?". * The answer is: "you prefix it with a `::`". */fun main(args: Array<String>) {    val numbers = listOf(1, 2, 3)    println(numbers.filter(::isOdd))}fun isOdd(x: Int) = x % 2 != 0

运行结果： [1, 3]

复合函数

看了下面的复合函数的例子，你会发现Kotlin的FP的实现相当简洁。（跟纯数学的表达式，相当接近了）

/** * The composition function return a composition of two functions passed to it: * compose(f, g) = f(g(*)). * Now, you can apply it to callable references. */fun main(args: Array<String>) {    val oddLength = compose(::isOdd, ::length)    val strings = listOf("a", "ab", "abc")    println(strings.filter(oddLength))}fun isOdd(x: Int) = x % 2 != 0fun length(s: String) = s.lengthfun <A, B, C> compose(f: (B) -> C, g: (A) -> B): (A) -> C {    return { x -> f(g(x)) }}

运行结果： [a,abc]

简单说明下

val oddLength = compose(::isOdd, ::length)    val strings = listOf("a", "ab", "abc")    println(strings.filter(oddLength))

这就是数学中，复合函数的定义：

h = h(f(g))

g: A->B
f: B->C
h: A->C

g(A)=B
h(A) = f(B) = f(g(A)) = C

只是代码中的写法是：

h=compose( f, g )
h=compose( f(g(A)), g(A) )

/** * The composition function return a composition of two functions passed to it: * compose(f, g) = f(g(*)). * Now, you can apply it to callable references. */fun main(args: Array<String>) {    val oddLength = compose(::isOdd, ::length)    val strings = listOf("a", "ab", "abc")    println(strings.filter(oddLength))        println(strings.filter(::hasA))    println(strings.filter(::hasB))        val hasBStrings = strings.filter(::hasB)    println(hasBStrings)        val evenLength = compose(::isEven,::length)    println(hasBStrings.filter(evenLength))        }fun isOdd(x: Int) = x % 2 != 0fun isEven(x:Int) = x % 2 == 0fun length(s: String) = s.lengthfun hasA(x: String) = x.contains("a")fun hasB(x: String) = x.contains("b")fun <A, B, C> compose(f: (B) -> C, g: (A) -> B): (A) -> C {    return { x -> f(g(x)) }}fun <W,X,Y,Z> compose2( h: (Y) -> Z, f:(X) -> Y,g:(W) -> X): (W) -> Z {    return  {x -> h(f(g(x)))} }

运行结果：

[a, abc][a, ab, abc][ab, abc][ab, abc][ab]