Go has confusing pointers

[chris...@gmail.com]

Hi,

Please help me understand the below:

I got this from "The Zoo of Go Functions"

type Count int

func (c *Count) Incr() int {

 *c = *c + 1 //why use *c and not just c, based on (c *Count) c is already a pointer?

 return int(*c) //why cast to int, c is already of type int? 

}

Christof

[Michael Jones]

this says "increment the data c points at" not "increment the pointer"

the function is returning an int instead of a Count 

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Michael T. Jones
michae...@gmail.com

[Tyler Compton]

 *c = *c + 1 //why use *c and not just c, based on (c *Count) c is already a pointer?

In Go (as well as other languages like C) '*' plays two roles. When used with a type like in "c *Count", it tells you that c is a pointer to a Count object. When used on a variable, like "*c + 1", it dereferences the pointer. In other words, "*c + 1" gets the value that "c" is pointing to and adds one to that value.

 

why cast to int, c is already of type int? 

In this case, c is of type Count, not int. When you define a type in Go, you're defining a new distinct type, not an alias to the underlying type. The Count type is distinct from int, even though it's based on int.

[howar...@gmail.com]

c is of type *Count (pointer to Count), surely? It is (*c) that is of type Count, which has to be cast to int to be returned as such because Go requires explicit casts. Agree with the rest.

[Tyler Compton]

c is of type *Count (pointer to Count), surely?

Yes, my mistake! Thank you for the correction.