Go中的一级函数

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英文:

First class functions in Go

问题

我来自JavaScript,它支持一级函数。例如,你可以:

  • 将一个函数作为参数传递给另一个函数
  • 从一个函数中返回一个函数。

有人可以给我一个在Go中如何实现这个的例子吗?

英文:

I come from JavaScript which has first class function support. For example you can:

  • pass a function as a parameter to another function
  • return a function from a function.

Can someone give me an example of how I would do this in Go?

答案1

得分: 42

Go语言和函数式编程可能会有所帮助。以下是这篇博客文章的内容:

package main
import fmt "fmt"
type Stringy func() string
func foo() string{
        return "Stringy function"
}
func takesAFunction(foo Stringy){
    fmt.Printf("takesAFunction: %v\n", foo())
}
func returnsAFunction()Stringy{
    return func()string{
        fmt.Printf("内部的stringy函数\n");
        return "bar" // 必须返回一个字符串才能是stringy类型
    }
}
func main(){
    takesAFunction(foo);
    var f Stringy = returnsAFunction();
    f();
    var baz Stringy = func()string{
        return "匿名的stringy\n"
    };
    fmt.Printf(baz());
}

作者是博客所有者:Dethe Elza(不是我)

英文:

Go Language and Functional Programming might help. From this blog post:

package main
import fmt "fmt"
type Stringy func() string
func foo() string{
        return "Stringy function"
}
func takesAFunction(foo Stringy){
    fmt.Printf("takesAFunction: %v\n", foo())
}
func returnsAFunction()Stringy{
    return func()string{
        fmt.Printf("Inner stringy function\n");
        return "bar" // have to return a string to be stringy
    }
}
func main(){
    takesAFunction(foo);
    var f Stringy = returnsAFunction();
    f();
    var baz Stringy = func()string{
        return "anonymous stringy\n"
    };
    fmt.Printf(baz());
}

Author is the blog owner: Dethe Elza (not me)

答案2

得分: 26

package main

import (
"fmt"
)

type Lx func(int) int

func cmb(f, g Lx) Lx {
return func(x int) int {
return g(f(x))
}
}

func inc(x int) int {
return x + 1
}

func sum(x int) int {
result := 0

for i := 0; i < x; i++ {
	result += i
}

return result

}

func main() {
n := 666

fmt.Println(cmb(inc, sum)(n))
fmt.Println(n * (n + 1) / 2)

}

英文:
package main

import (
	&quot;fmt&quot;
)

type Lx func(int) int

func cmb(f, g Lx) Lx {
	return func(x int) int {
		return g(f(x))
	}
}

func inc(x int) int {
	return x + 1
}

func sum(x int) int {
	result := 0

	for i := 0; i &lt; x; i++ {
		result += i
	}

	return result
}

func main() {
	n := 666

	fmt.Println(cmb(inc, sum)(n))
	fmt.Println(n * (n + 1) / 2)
}

output:

222111
222111

答案3

得分: 6

从规范中提取的相关部分:函数类型

这里的其他答案首先声明了一个新类型,这是很好的做法,可以使代码更易读,但要知道这不是必需的。

您可以在不声明新类型的情况下使用函数值,如下面的示例所示。

声明一个具有两个float64类型参数和一个返回值为float64类型的函数类型的变量如下所示:

// 创建一个具有上述函数类型的变量:
var f func(float64, float64) float64

让我们编写一个返回加法函数的函数。这个加法函数应该接受两个float64类型的参数,并在调用时返回这两个数字的和:

func CreateAdder() func(float64, float64) float64 {
    return func(x, y float64) float64 {
        return x + y
    }
}

让我们编写一个具有3个参数的函数,前两个参数是float64类型,第三个参数是一个函数值,一个接受两个float64类型输入参数并产生float64类型值的函数。我们正在编写的函数将调用作为参数传递给它的函数值,并使用前两个float64值作为函数值的参数,并返回传递的函数值返回的结果:

func Execute(a, b float64, op func(float64, float64) float64) float64 {
    return op(a, b)
}

让我们看看我们之前的示例的运行情况:

var adder func(float64, float64) float64 = CreateAdder()
result := Execute(1.5, 2.5, adder)
fmt.Println(result) // 输出 4

请注意,当创建adder时,您当然可以使用短变量声明

adder := CreateAdder() // adder 的类型为:func(float64, float64) float64

Go Playground上尝试这些示例。

使用现有函数

当然,如果您已经在一个包中声明了具有相同函数类型的函数,您也可以使用它。

例如,math.Mod()具有相同的函数类型:

func Mod(x, y float64) float64

因此,您可以将此值传递给我们的Execute()函数:

fmt.Println(Execute(12, 10, math.Mod)) // 输出 2

输出为2,因为12 mod 10 = 2。请注意,现有函数的名称充当函数值。

Go Playground上尝试它。

注意:

请注意,参数名称不是类型的一部分,具有相同参数和结果类型的两个函数的类型是相同的,而不管参数的名称如何。但要知道,在参数或结果列表中,参数的名称必须全部存在或全部不存在。

因此,例如,您也可以编写:

func CreateAdder() func(P float64, Q float64) float64 {
    return func(x, y float64) float64 {
        return x + y
    }
}

或者:

var adder func(x1, x2 float64) float64 = CreateAdder()
英文:

The related section from the specification: Function types.

All other answers here first declare a new type, which is good (practice) and makes your code easier to read, but know that this is not a requirement.

You can work with function values without declaring a new type for them, as seen in the below example.

Declaring a variable of function type which has 2 parameters of type float64 and has one return value of type float64 looks like this:

// Create a var of the mentioned function type:
var f func(float64, float64) float64

Let's write a function which returns an adder function. This adder function should take 2 parameters of type float64 and should returns the sum of those 2 numbers when called:

func CreateAdder() func(float64, float64) float64 {
    return func(x, y float64) float64 {
        return x + y
    }
}

Let's write a function which has 3 parameters, first 2 being of type float64, and the 3rd being a function value, a function that takes 2 input parameters of type float64 and produces a value of float64 type. And the function we're writing will call the function value that is passed to it as parameter, and using the first 2 float64 values as arguments for the function value, and returns the result that the passed function value returns:

func Execute(a, b float64, op func(float64, float64) float64) float64 {
    return op(a, b)
}

Let's see our previous examples in action:

var adder func(float64, float64) float64 = CreateAdder()
result := Execute(1.5, 2.5, adder)
fmt.Println(result) // Prints 4

Note that of course you can use the Short variable declaration when creating adder:

adder := CreateAdder() // adder is of type: func(float64, float64) float64

Try these examples on the Go Playground.

Using an existing function

Of course if you already have a function declared in a package with the same function type, you can use that too.

For example the math.Mod() has the same function type:

func Mod(x, y float64) float64

So you can pass this value to our Execute() function:

fmt.Println(Execute(12, 10, math.Mod)) // Prints 2

Prints 2 because 12 mod 10 = 2. Note that the name of an existing function acts as a function value.

Try it on the Go Playground.

Note:

Note that the parameter names are not part of the type, the type of 2 functions having the same parameter and result types is identical regardless of the names of the parameters. But know that within a list of parameters or results, the names must either all be present or all be absent.

So for example you can also write:

func CreateAdder() func(P float64, Q float64) float64 {
    return func(x, y float64) float64 {
        return x + y
    }
}

Or:

var adder func(x1, x2 float64) float64 = CreateAdder()

答案4

得分: 1

虽然你可以使用var或声明类型,但你不需要这样做。
你可以很简单地这样做:

package main

import "fmt"

var count int

func increment(i int) int {
    return i + 1
}

func decrement(i int) int {
    return i - 1
}

func execute(f func(int) int) int {
    return f(count)
}

func main() {
    count = 2
    count = execute(increment)
    fmt.Println(count)
    count = execute(decrement)
    fmt.Println(count)
}

//输出结果为:
3
2
英文:

While you can use a var or declare a type, you don't need to.
You can do this quite simply:

package main

import &quot;fmt&quot;

var count int

func increment(i int) int {
	return i + 1
}

func decrement(i int) int {
	return i - 1
}

func execute(f func(int) int) int {
	return f(count)
}

func main() {
	count = 2
	count = execute(increment)
	fmt.Println(count)
	count = execute(decrement)
	fmt.Println(count)
}

//The output is:
3
2

答案5

得分: 0

只是一个使用递归函数定义的脑筋急转弯,用于在Web应用程序中链接中间件。

首先,是工具箱:

func MakeChain() (Chain, http.Handler) {
    nop := http.HandlerFunc(func(res http.ResponseWriter, req *http.Request) {})
    var list []Middleware
    var final http.Handler = nop
    var f Chain
    f = func(m Middleware) Chain {
        if m != nil {
            list = append(list, m)
        } else {
            for i := len(list) - 1; i >= 0; i-- {
                mid := list[i]

                if mid == nil {
                    continue
                }

                if next := mid(final); next != nil {
                    final = next
                } else {
                    final = nop
                }
            }

            if final == nil {
                final = nop
            }
            return nil
        }
        return f
    }
    return f, final
}

type (
    Middleware func(http.Handler) http.Handler
    Chain      func(Middleware) Chain
)

如你所见,类型Chain是一个返回相同类型Chain的另一个函数(多么一流的类型!)。

现在进行一些测试以查看其运行情况:

func TestDummy(t *testing.T) {
    c, final := MakeChain()
    c(mw1(`OK!`))(mw2(t, `OK!`))(nil)
    log.Println(final)

    w1 := httptest.NewRecorder()
    r1, err := http.NewRequest("GET", "/api/v1", nil)
    if err != nil {
        t.Fatal(err)
    }
    final.ServeHTTP(w1, r1)
}

func mw2(t *testing.T, expectedState string) func(next http.Handler) http.Handler {
    return func(next http.Handler) http.Handler {
        return http.HandlerFunc(func(w http.ResponseWriter, r *http.Request) {
            val := r.Context().Value(contextKey("state"))
            sval := fmt.Sprintf("%v", val)
            assert.Equal(t, sval, expectedState)
        })
    }
}

func mw1(initialState string) func(next http.Handler) http.Handler {
    return func(next http.Handler) http.Handler {
        return http.HandlerFunc(func(w http.ResponseWriter, r *http.Request) {
            ctx := context.WithValue(r.Context(), contextKey("state"), initialState)
            next.ServeHTTP(w, r.WithContext(ctx))
        })
    }
}

type contextKey string

再次强调,这只是一个脑筋急转弯,展示了我们可以在Go中以不同的方式使用一流函数。个人而言,我现在使用chi作为路由器和处理中间件。

英文:

Just a brainteaser with recursive function definition for chaining middlewares in a web app.

First, the toolbox:

func MakeChain() (Chain, http.Handler) {
	nop := http.HandlerFunc(func(res http.ResponseWriter, req *http.Request) {})
	var list []Middleware
	var final http.Handler = nop
	var f Chain
	f = func(m Middleware) Chain {
		if m != nil {
			list = append(list, m)
		} else {
			for i := len(list) - 1; i &gt;= 0; i-- {
				mid := list[i]

				if mid == nil {
					continue
				}

				if next := mid(final); next != nil {
					final = next
				} else {
					final = nop
				}
			}

			if final == nil {
				final = nop
			}
			return nil
		}
		return f
	}
	return f, final
}

type (
	Middleware func(http.Handler) http.Handler
	Chain      func(Middleware) Chain
)

As you see type Chain is a function that returns another function of the same type Chain (How first class is that!).

Now some tests to see it in action:

func TestDummy(t *testing.T) {
	c, final := MakeChain()
	c(mw1(`OK!`))(mw2(t, `OK!`))(nil)
	log.Println(final)

	w1 := httptest.NewRecorder()
	r1, err := http.NewRequest(&quot;GET&quot;, &quot;/api/v1&quot;, nil)
	if err != nil {
		t.Fatal(err)
	}
	final.ServeHTTP(w1, r1)
}

func mw2(t *testing.T, expectedState string) func(next http.Handler) http.Handler {
	return func(next http.Handler) http.Handler {
		return http.HandlerFunc(func(w http.ResponseWriter, r *http.Request) {
			val := r.Context().Value(contextKey(&quot;state&quot;))
			sval := fmt.Sprintf(&quot;%v&quot;, val)
			assert.Equal(t, sval, expectedState)
		})
	}
}

func mw1(initialState string) func(next http.Handler) http.Handler {
	return func(next http.Handler) http.Handler {
		return http.HandlerFunc(func(w http.ResponseWriter, r *http.Request) {
			ctx := context.WithValue(r.Context(), contextKey(&quot;state&quot;), initialState)
			next.ServeHTTP(w, r.WithContext(ctx))
		})
	}
}

type contextKey string

Again, this was just a brainteaser to show we can use first class functions in Go in different ways. Personally I use chi nowadays as router and for handling middlewares.

huangapple
  • 本文由 发表于 2010年12月5日 16:45:41
  • 转载请务必保留本文链接:https://go.coder-hub.com/4358031.html
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