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11.14 结构体、集合和高阶函数

通常你在应用中定义了一个结构体，那么你也可能需要这个结构体的（指针）对象集合，比如：

type Any interface{}
type Car struct {
	Model        string
	Manufacturer string
	BuildYear    int
	// ...
}

type Cars []*Car

然后我们就可以使用高阶函数，实际上也就是把函数作为定义所需方法（其他函数）的参数，例如：

1）定义一个通用的 Process() 函数，它接收一个作用于每一辆 car 的 f 函数作参数：

// Process all cars with the given function f:
func (cs Cars) Process(f func(car *Car)) {
	for _, c := range cs {
		f(c)
	}
}

2）在上面的基础上，实现一个查找函数来获取子集合，并在 Process() 中传入一个闭包执行（这样就可以访问局部切片 cars）：

// Find all cars matching a given criteria.
func (cs Cars) FindAll(f func(car *Car) bool) Cars {

	cars := make([]*Car, 0)
	cs.Process(func(c *Car) {
		if f(c) {
			cars = append(cars, c)
		}
	})
	return cars
}

3）实现对应作用的功效 (Map-functionality)，从每个 car 对象当中产出某些东西：

// Process cars and create new data.
func (cs Cars) Map(f func(car *Car) Any) []Any {
	result := make([]Any, 0)
	ix := 0
	cs.Process(func(c *Car) {
		result[ix] = f(c)
		ix++
	})
	return result
}

现在我们可以定义下面这样的具体查询：

allNewBMWs := allCars.FindAll(func(car *Car) bool {
	return (car.Manufacturer == "BMW") && (car.BuildYear > 2010)
})

4）我们也可以根据参数返回不同的函数。也许我们想根据不同的厂商添加汽车到不同的集合，但是这（这种映射关系）可能会是会改变的。所以我们可以定义一个函数来产生特定的添加函数和 map 集：

func MakeSortedAppender(manufacturers []string)(func(car *Car),map[string]Cars) {
	// Prepare maps of sorted cars.
	sortedCars := make(map[string]Cars)
	for _, m := range manufacturers {
		sortedCars[m] = make([]*Car, 0)
	}
	sortedCars["Default"] = make([]*Car, 0)
	// Prepare appender function:
	appender := func(c *Car) {
		if _, ok := sortedCars[c.Manufacturer]; ok {
			sortedCars[c.Manufacturer] = append(sortedCars[c.Manufacturer], c)
		} else {
			sortedCars["Default"] = append(sortedCars["Default"], c)
		}

	}
	return appender, sortedCars
}

现在我们可以用它把汽车分类为独立的集合，像这样：

manufacturers := []string{"Ford", "Aston Martin", "Land Rover", "BMW", "Jaguar"}
sortedAppender, sortedCars := MakeSortedAppender(manufacturers)
allUnsortedCars.Process(sortedAppender)
BMWCount := len(sortedCars["BMW"])

我们让这些代码在下面的程序 cars.go 中执行：

示例 11.18 [cars.go]：

// cars.go
package main

import (
	"fmt"
)

type Any interface{}
type Car struct {
	Model        string
	Manufacturer string
	BuildYear    int
	// ...
}
type Cars []*Car

func main() {
	// make some cars:
	ford := &Car{"Fiesta", "Ford", 2008}
	bmw := &Car{"XL 450", "BMW", 2011}
	merc := &Car{"D600", "Mercedes", 2009}
	bmw2 := &Car{"X 800", "BMW", 2008}
	// query:
	allCars := Cars([]*Car{ford, bmw, merc, bmw2})
	allNewBMWs := allCars.FindAll(func(car *Car) bool {
		return (car.Manufacturer == "BMW") && (car.BuildYear > 2010)
	})
	fmt.Println("AllCars: ", allCars)
	fmt.Println("New BMWs: ", allNewBMWs)
	//
	manufacturers := []string{"Ford", "Aston Martin", "Land Rover", "BMW", "Jaguar"}
	sortedAppender, sortedCars := MakeSortedAppender(manufacturers)
	allCars.Process(sortedAppender)
	fmt.Println("Map sortedCars: ", sortedCars)
	BMWCount := len(sortedCars["BMW"])
	fmt.Println("We have ", BMWCount, " BMWs")
}

// Process all cars with the given function f:
func (cs Cars) Process(f func(car *Car)) {
	for _, c := range cs {
		f(c)
	}
}

// Find all cars matching a given criteria.
func (cs Cars) FindAll(f func(car *Car) bool) Cars {
	cars := make([]*Car, 0)

	cs.Process(func(c *Car) {
		if f(c) {
			cars = append(cars, c)
		}
	})
	return cars
}

// Process cars and create new data.
func (cs Cars) Map(f func(car *Car) Any) []Any {
	result := make([]Any, len(cs))
	ix := 0
	cs.Process(func(c *Car) {
		result[ix] = f(c)
		ix++
	})
	return result
}

func MakeSortedAppender(manufacturers []string) (func(car *Car), map[string]Cars) {
	// Prepare maps of sorted cars.
	sortedCars := make(map[string]Cars)

	for _, m := range manufacturers {
		sortedCars[m] = make([]*Car, 0)
	}
	sortedCars["Default"] = make([]*Car, 0)

	// Prepare appender function:
	appender := func(c *Car) {
		if _, ok := sortedCars[c.Manufacturer]; ok {
			sortedCars[c.Manufacturer] = append(sortedCars[c.Manufacturer], c)
		} else {
			sortedCars["Default"] = append(sortedCars["Default"], c)
		}
	}
	return appender, sortedCars
}

输出：

AllCars:  [0xf8400038a0 0xf840003bd0 0xf840003ba0 0xf840003b70]
New BMWs:  [0xf840003bd0]
Map sortedCars:  map[Default:[0xf840003ba0] Jaguar:[] Land Rover:[] BMW:[0xf840003bd0 0xf840003b70] Aston Martin:[] Ford:[0xf8400038a0]]
We have  2  BMWs

devTest.run

11.14 结构体、集合和高阶函数

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