To initialize a map in a struct one should do the following:
someStruct.nestedMap = make(map[int8]int8)
But what should you do if you have a code structure like this:
type Base struct {
base map[int8]uint64
}
type Middle struct {
baseObjects map[int8]Base
}
type Top struct {
middleObjects map[int8]Middle
}
There we have a total of 3 structs where each has a struct as a key. How do you initialize this, and make it ready to go?
要在结构中初始化映射，应执行以下操作：
someStruct.nestedMap = make（map [int8] int8）
但是，如果您具有这样的代码结构，该怎么办：
type基础结构{
基础结构图[int8] uint64
}
type中间结构{
baseObjects map [int8] Base
}
type顶部结构{
middleObjects map [int8] Middle
}
共有3个结构，每个结构都有一个键作为结构体。
如何初始化 并准备好了吗？
What you want is auctually a map with default value that is not zero value, but a value ready to use, in this case, a made map.
Before going to the solution, there auctually is an issue about go here: https://github.com/golang/go/issues/3117
For reasons, Go now does not support to assign to a field of a struct if the struct is stored in a map.
To solve the issue, you need to use a pointer to that struct, and store the pointer in the map instead. So your data structure needs to be changed like:
微信扫一扫type Base struct {
base map[int8]uint64
}
type Middle struct {
baseObjects map[int8]*Base
}
type Top struct {
middleObjects map[int8]*Middle
}
And to the core logic: a defualt value map. Go does not support that with plain maps, but we can extend it, using methods to wrap it around.
func (t Top) Get(i int8) *Middle {
x, ok := t.middleObjects[i]
if !ok {
v := NewMiddle()
t.middleObjects[i] = v
return v
}
return x
}
func (m Middle) Get(i int8) *Base {
x, ok := m.baseObjects[i]
if !ok {
v := NewBase()
m.baseObjects[i] = v
return v
}
return x
}
When we are trying to get a value, we first check if it exists. If not, we return a new constructed one, and if it did exist, return the value.
Usage:
t.Get(8).Get(9).base[10] = 14
Playground example: https://play.golang.org/p/0JSN0yjRPif