Reverse 在 Golang 的底层是如何工作的？

Question

s := []string{"Zeno", "John", "Al", "Jenny"}

sort.Sort(sort.Reverse(sort.StringSlice(s)))

我无法理解的逻辑Reverse

Reverse 的源代码如下所示：

func Reverse(data Interface) Interface {

    return &reverse{data}

}

type reverse struct {

    // This embedded Interface permits Reverse to use the methods of

    // another Interface implementation.

    Interface

}

type Interface interface {

    // Len is the number of elements in the collection.

    Len() int

    // Less reports whether the element with index i

    // must sort before the element with index j.

    //

    // If both Less(i, j) and Less(j, i) are false,

    // then the elements at index i and j are considered equal.

    // Sort may place equal elements in any order in the final result,

    // while Stable preserves the original input order of equal elements.

    //

    // Less must describe a transitive ordering:

    //  - if both Less(i, j) and Less(j, k) are true, then Less(i, k) must be true as well.

    //  - if both Less(i, j) and Less(j, k) are false, then Less(i, k) must be false as well.

    //

    // Note that floating-point comparison (the < operator on float32 or float64 values)

    // is not a transitive ordering when not-a-number (NaN) values are involved.

    // See Float64Slice.Less for a correct implementation for floating-point values.

    Less(i, j int) bool

    // Swap swaps the elements with indexes i and j.

    Swap(i, j int)

}

上面那些给定的操作如何使给定的数组反转？

Answer 1

如果你看一下 sort.StringSlice 类型，你可以看到它实现了 Less 方法，注意比较 x[i] < x[j]，这意味着较小的元素在前。func (x StringSlice) Less(i, j int) bool { return x[i] < x[j] }然后注意 sort.reverse 类型（不是 sort.Reverse 接口），它也实现了 Less 方法，但看看它如何传递 i 和 j 参数，接收 i 和 j 但传递 j 和 i，这相当于 x[ i] > x[j]// Less returns the opposite of the embedded implementation's Less method.func (r reverse) Less(i, j int) bool {&nbsp; &nbsp; return r.Interface.Less(j, i)}