在Java中,我可以这样做
derp(new Runnable { public void run () { /* run this sometime later */ } })
然后在方法中“运行”代码。处理起来很麻烦(匿名内部类),但这是可以做到的。
Go是否有一些可以方便函数/回调作为参数传递的东西?
在Java中,我可以这样做
derp(new Runnable { public void run () { /* run this sometime later */ } })
然后在方法中“运行”代码。处理起来很麻烦(匿名内部类),但这是可以做到的。
Go是否有一些可以方便函数/回调作为参数传递的东西?
是的,考虑一下这些例子:
package main
import "fmt"
// convert types take an int and return a string value.
type convert func(int) string
// value implements convert, returning x as string.
func value(x int) string {
return fmt.Sprintf("%v", x)
}
// quote123 passes 123 to convert func and returns quoted string.
func quote123(fn convert) string {
return fmt.Sprintf("%q", fn(123))
}
func main() {
var result string
result = value(123)
fmt.Println(result)
// Output: 123
result = quote123(value)
fmt.Println(result)
// Output: "123"
result = quote123(func(x int) string { return fmt.Sprintf("%b", x) })
fmt.Println(result)
// Output: "1111011"
foo := func(x int) string { return "foo" }
result = quote123(foo)
fmt.Println(result)
// Output: "foo"
_ = convert(foo) // confirm foo satisfies convert at runtime
// fails due to argument type
// _ = convert(func(x float64) string { return "" })
}
玩:http://play.golang.org/p/XNMtrDUDS0
参观:https://tour.golang.org/moretypes/25(函数闭包)
这里有一个简单的例子:
package main
import "fmt"
func plusTwo() (func(v int) (int)) {
return func(v int) (int) {
return v+2
}
}
func plusX(x int) (func(v int) (int)) {
return func(v int) (int) {
return v+x
}
}
func main() {
p := plusTwo()
fmt.Printf("3+2: %d\n", p(3))
px := plusX(3)
fmt.Printf("3+3: %d\n", px(3))
}
你可以把函数作为参数传递给Go函数。下面是一个将函数作为参数传递给另一个Go函数的例子:
package main
import "fmt"
type fn func(int)
func myfn1(i int) {
fmt.Printf("\ni is %v", i)
}
func myfn2(i int) {
fmt.Printf("\ni is %v", i)
}
func test(f fn, val int) {
f(val)
}
func main() {
test(myfn1, 123)
test(myfn2, 321)
}
你可以在https://play.golang.org/p/9mAOUWGp0k上尝试一下
下面是Go中的“Map”实现示例。希望这能有所帮助!!
func square(num int) int {
return num * num
}
func mapper(f func(int) int, alist []int) []int {
var a = make([]int, len(alist), len(alist))
for index, val := range alist {
a[index] = f(val)
}
return a
}
func main() {
alist := []int{4, 5, 6, 7}
result := mapper(square, alist)
fmt.Println(result)
}
我希望下面的例子能更清楚地说明问题。
package main
type EmployeeManager struct{
category string
city string
calculateSalary func() int64
}
func NewEmployeeManager() (*EmployeeManager,error){
return &EmployeeManager{
category : "MANAGEMENT",
city : "NY",
calculateSalary: func() int64 {
var calculatedSalary int64
// some formula
return calculatedSalary
},
},nil
}
func (self *EmployeeManager) emWithSalaryCalculation(){
self.calculateSalary = func() int64 {
var calculatedSalary int64
// some new formula
return calculatedSalary
}
}
func updateEmployeeInfo(em EmployeeManager){
// Some code
}
func processEmployee(){
updateEmployeeInfo(struct {
category string
city string
calculateSalary func() int64
}{category: "", city: "", calculateSalary: func() int64 {
var calculatedSalary int64
// some new formula
return calculatedSalary
}})
}
你也可以传递一个struct的函数,比如:
package main
// define struct
type Apple struct {}
// return apple's color
func (Apple) GetColor() string {
return "Red"
}
func main () {
// instantiate
myApple := Apple{}
// put the func in a variable
theFunc := myApple.GetColor
// execute the variable as a function
color := theFunc()
print(color)
}
输出将为“红色”,检查操场
这是我能想到的最简单的方法。
package main
import "fmt"
func main() {
g := greeting
getFunc(g)
}
func getFunc(f func()) {
f()
}
func greeting() {
fmt.Println("Hello")
}