Go是否允许函数将另一个函数用作参数？

Question

问题发生在Go代码的第17行.下面是python和Go中的程序,所以你可以看到我正在尝试做什么.Python工作,我的Go尝试都失败了.已经连续阅读了golang.org,谷歌也没有看到任何内容.

def my_filter(x):
  if x % 5 == 0:
    return True
  return False

#Function which returns a list of those numbers which satisfy the filter
def my_finc(Z, my_filter):

  a = []
  for x in Z:
    if my_filter(x) == True:
      a.append(x)
  return a

print(my_finc([10, 4, 5, 17, 25, 57, 335], my_filter))

现在,Go版本我遇到了麻烦:

package main

import "fmt"

func Filter(a []int) bool {
    var z bool
    for i := 0; i < len(a); i++ {
        if a[i]%5 == 0 {
            z = true
        } else {
            z = false
        }
    }
    return z
}

func finc(b []int, Filter) []int {
    var c []int
    for i := 0; i < len(c); i++ {
        if Filter(b) == true {
            c = append(c, b[i])
        }
    }
    return c
}

func main() {
    fmt.Println(finc([]int{1, 10, 2, 5, 36, 25, 123}, Filter))
}

Answer 1

是的,Go可以将函数作为参数:

package main

import "fmt"

func myFilter(a int) bool {
    return a%5 == 0
}

type Filter func(int) bool

func finc(b []int, filter Filter) []int {
    var c []int
    for _, i := range b {
        if filter(i) {
            c = append(c, i)
        }
    }
    return c
}

func main() {
    fmt.Println(finc([]int{1, 10, 2, 5, 36, 25, 123}, myFilter))
}

关键是你需要传递一种类型.

type Filter func(int) bool

我还清理了一些代码,使其更具惯用性.我用范围子句替换了你的for循环.

for i := 0; i < len(b); i++ {
    if filter(b[i]) == true {
        c = append(c, b[i])
    }
}

变

for _, i := range b {
    if filter(i) {
        c = append(c, i)
    }
}