Rule for Go Pointers, References, Dereferencing:

Question

I am new to GoLang, coming from the Delphi, C++ world - admittedly very excited about this language, which I think is destined to become \"the next big thing\".

I am

Answer 1

All of the methods for a List have *List receivers: (http://golang.org/pkg/container/list/)

func (l *List) Back() *Element
func (l *List) Front() *Element
func (l *List) Init() *List
...
func (l *List) Remove(e *Element) interface{}

In your example l is of type *List, so there's no need to dereference them.

Suppose, instead, that you had something like this:

type A struct{}
func (a  A) X() {
    fmt.Println("X")
}
func (a *A) Y() {
    fmt.Println("Y")
}

You are allowed to write:

a := A{}
a.X()
a.Y() // == (&a).Y()

Or you can do the following:

a := &A{}
a.X() // same like == (*a).X()
a.Y()

But it only works for method receivers. Go will not automatically convert function arguments. Given these functions:

func A(x *int) {
    fmt.Println(*x)
}
func B(y int) {
    fmt.Println(y)
}

This is invalid:

A(5)

You have to do this:

var x int 
A(&x)

This is also invalid:

var y *int
B(y)

You have to do this:

B(*y)

Unlike C# or Java, when it comes to structs, Go does not make a distinction between reference and value types. A *List is a pointer, a List is not. Modifying a field on a List only modifies the local copy. Modifying a field on a *List modifies all "copies". (cause they aren't copies... they all point to the same thing in memory)

There are types which seem to hide the underlying pointer (like a slice contains a pointer to an array), but Go is always pass by value.