An interview question about iota in golang

What is the output of the following program？

package main

import "fmt"

const (
	a = iota
	b = iota
)
const (
	name = "xxxx"
	c    = iota
	d    = iota
)

const a0 = iota 

const (
    a1 = iota   
    a2 = iota  
    a3 = 6      
    a4          
    a5 = iota  
)

func main() {
    
	fmt.Println(a)
	fmt.Println(b)
	fmt.Println(c)
	fmt.Println(d)

	fmt.Println(a1)
	fmt.Println(a2)
	fmt.Println(a3)
	fmt.Println(a4)
	fmt.Println(a5)
}

请思考,答案在下文.....
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0
1
1
2
0
1
6
6
4

iota使用的规则如下:

• 每当const出现时, 都会使iota初始化为0.
• const中每新增一行常量声明将使iota计数一次.

const (
    a1 = iota   // a1 = 0 // 又一个const出现, iota初始化为0
    a2 = iota   // a1 = 1 // const新增一行, iota 加1
    a3 = 6      // a3 = 6 // define a constant
    a4          // a4 = 6 // No assignment is the same as the previous line
    a5 = iota   // a5 = 4 // const已经新增了4行, 所以这里是4
)