Go pointer dereferencing

Question

I'm currently trying to learn GO and mainly knowing and working with Java, ASP.Net and some Python, there is no experience working with C-like pointers, which causes my current confusion.

A library I'm currently using to write my first GO project is called Commando. There I have the struct CommandRegistry and the variable of interest is called Commands.

In the struct the variable is described as the following:

// registered command configurations
Commands map[string]*Command

On a first glimpse I would understand this as a Map object containing a list of Strings, however it also shows the pointer reference to the actual Command object.

All I can see is that it is a map I can loop over which returns the name of the command ( the string ), however I'm wondering if the *Command in the type description means I can somehow dereference the pointer and retrieve the object itself to extract the additional information of it.

As I know the & operand is used to create a new pointer of another object. Pass-by-reference basically instead of pass-by-value. And the * operand generally signals the object is a pointer or used to require a pointer in a new function.

Is there a way I can retrieve the Command object or why does the type contain the *Command in it's declaration?

it's a map, so you can get the value for your string key Commands["key"] should give you the pointer to your Command. In Java this would be declared as Map<String, Command> - without pointers. — Thomas Jungblut
– Thomas Jungblut, Commented Oct 21, 2021 at 10:37
Commands is a map, it maps from string values to *Command, that is, pointer to Command. Indexing the map will give you the associated *Command pointer, e.g. Command["foo"]. You can dereference that (if it's not nil). You can also iterate over the map with a loop, getting all key-value pairs, like this: for s, cmd := range Commands { fmt.Printf("key: %s, value: %v", s, cmd) } — icza
– icza, Commented Oct 21, 2021 at 10:38
Thanks a lot for the quick comments. Indeed it seems I was blinded by the new topic of pointers and overlooked the simple solution that it still is a map just declared differently. — Nico
– Nico, Commented Oct 21, 2021 at 10:41
Note that you don't have to dereference the pointer in most cases, you can access the struct's fields and call it's methods using the pointer, it is basically short-hard for dereferencing so that you don't have to do it every time. — mkopriva
– mkopriva, Commented Oct 21, 2021 at 10:41

Thijs van der Heijden · Accepted Answer · 2021-10-21 10:41:03Z

2

Commands is a map (dictionary) which has strings as keys, and pointers to Commands as values. By passing it a key, you will get a pointer to the command it belongs to. You can then dereference the pointer to an actual Command object by using the * operator. Something like dereferencedCommand := *Commands["key"].

The * operator can be quite confusing, at least it was for me. When used as a type it denotes that we are receiving the memory address of some variable. But to dereference a memory address to a concrete type, you also use the * operator.

answered Oct 21, 2021 at 10:41

Thijs van der Heijden

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1 Comment

Nico Over a year ago

Thanks a lot for the answer and explanation of the * operator. It indees is a bit confusing when working with it for the first type.

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