以编程方式创建嵌套数组

问题描述：

I want to test my serialization format for robustnes and want to serialize a nested array. The Serialization actually dies at some point, but I don't know at what depth exactly. Right now I just define a large array manually, but I want to do it programatically, so I can figure out when exactly the depth is too high. Changing the code for each iteration is tedious...

l := [][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][][]string {{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{"a"}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}

What I want is something like this:

l := "a"
while(1){
    l = {l} //nest the slice/array once more
    print("round ", i)
    i++
}

I'm a Golang newbie and I don't know how I can get around the type-safety in this case.

我想测试我的序列化格式的鲁棒性，并希望序列化嵌套数组。序列化实际上在某个时候终止，但是我不知道确切的深度。现在，我只是手动定义一个大数组，但是我想以编程方式进行操作，因此我可以弄清楚深度到底何时过高。为每次迭代更改代码很繁琐... p>

  l：= [] [] [] [] [] [] [] [] [] [] [] [] []  [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] []  [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] []  [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] []  [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] [] string {{{{{{{{{{{{{{{{{{{{  {{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{  {{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{“一种”}}}}}}}}}}}}}}}  }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}  }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}} 
 代码>  PRE> 
 
  我想要的东西是这样的： p> 
 
 
  l：=“ a” 
while（1）{
l = {l} // //嵌套切片/数组 再一次
 print（“ round”，i）
 i ++ 
} 
  code>  pre> 
 
 我是Golang新手，我不知道该怎么办 在这种情况下避免类型安全。   p> 
  div>

答

In go you can't change the type of a variable so you need to use interface{} as type var l interface{} now you could do something like:

l = "a"
for i:= 0;i<3;i++{
    l = []interface{}{l}
}

printing l would give the output [[[a]]], same as [][][]string{{{"a"}}}.

A problem is that you lose your type and it will be impossible to do something like l[0][0][0] to get the value. As far as I now you would need to cast everything l.([]interface{})[0].([]interface{})[0].([]interface{})[0].(string)

以编程方式创建嵌套数组

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