使用闭包在Go中编写下一个置换，我的代码有什么问题

Question

I have written a function "iterPermutation" which uses closure. I want to return array and boolean from the closure which I could not do. So tried only array but it still gives an error

cannot use func literal (type func() []int) as type []int in return argument

I want to use iterPermutation like

a := []int{0,1,2,3,4}
nextPermutation, exists := iterPermutation(a)
for exists {
    nextPermutation()
}

func iterPermutation(a []int) []int {
    return func() []int {
        i := len(a) - 2

        for i >= 0 && a[i+1] <= a[i] {
            i--
        }
        if i < 0 {
            return a
        }

        j := len(a) - 1
        for j >= 0 && a[j] <= a[i] {
            j--
        }

        a[i], a[j] = a[j], a[i]

        for k, l := i, len(a)-1; k < l; k, l = k+1, l-1 {
            a[k], a[l] = a[l], a[k]
        }
        return a
    }
}

Answer 1

Golang spec for Return statements described:

The return value or values may be explicitly listed in the "return" statement. Each expression must be single-valued and assignable to the corresponding element of the function's result type.

The function called for permutation should contains two values in return one for the array and another for the boolean. Since you are assigning two variables from the function return:

a := []int{0,1,2,3,4}
nextPermutation, exists := iterPermutation(a) // it should return two values one for nextPermutation which is an array and other is exists which might be a boolean value.
for exists {
    nextPermutation()
}

For below error:

"cannot use func literal (type func() []int) as type []int in return argument"

you are returning func() literal enclosed inside the closure function of permutation along with boolean value, so change the return type as:

package main

func main(){
    a := []int{0,1,2,3,4}
    nextPermutation, _ := iterPermutation(a)
        nextPermutation()
}

func iterPermutation(a []int) ((func() []int), bool) { // return both values
    return func() []int {
        i := len(a) - 2

        for i >= 0 && a[i+1] <= a[i] {
            i--
        }
        if i < 0 {
            return a
        }

        j := len(a) - 1
        for j >= 0 && a[j] <= a[i] {
            j--
        }

        a[i], a[j] = a[j], a[i]

        for k, l := i, len(a)-1; k < l; k, l = k+1, l-1 {
            a[k], a[l] = a[l], a[k]
        }
        return a
    }, true // add boolean value to return from the function.
}

Working answer on Playground