Go函数具有返回/签名混合类型

Question

I'm trying to figure out how pointers work in Go and I think I'm starting to get it, but this is confusing me and I don't really know what to search for. Let's say I have the following function:

func createNode(nodeInfo string) *TreeNode {
    return &TreeNode{info: nodeInfo}
}

I understand that the function is returning the memory address of the created struct instance, but how does the function signature say *TreeNode? According to my understanding, the * is used to dereference pointers to get the value itself, so what is happening here?

Also, here:

 func zero(xPtr *int) {
  *xPtr = 0
}

func main() {
  x := 5
  zero(&x)
}

The opposite is happening where the function is accepting an argument with the * operator but the function itself is being called with & operator.

Answer 1

The * has 2 uses, one for variables and one for types:

• For types, it signifies, that the type is a pointer, not a value directly.
• For variables, it dereferences a pointer, as you already know. (One might also distinguish this use into to two "sub-uses", as * on the left side of assignments, as in *ptr = val sets the value the pointer is pointing to, while * usually "retrieves" the value the pointer is pointing to.)

& on the other hand, can only be used for variables and gets the address of an object in memory.

In your examples, the return type *TreeNode and the argument type *int signify that you are returning/expecting a pointer, according to the use for types. In contrast, *xPtr = 0 dereferences the variable xPtr.

To know which use is the correct one in your situation, you must make clear to yourself whether you are dealing with a type or a variable.

For a technical description, you can read the sections of the language specifications on Pointer types and Address operators.