Casting from one pointer to pointer type to another in Golang error

Question

Can anyone tell my why this wouldn't compile?

package main

type myint int
func set(a **myint) {
    i := myint(5)
    *a = &i 
}

func main() {
    var k *int
    set( (**myint)(&k) ) // cannot convert &k (type **int) to type **myint
    print( *k )
}

My reasoning so far is this. All types in Golang are different, but it allows to convert from one type to another with C-like cast syntax as long as underlying types are identical. In my example, converting 'int' to 'myint' is not a problem. '*int' to '*myint' isn't either. It's when you have pointer to pointer problems arise. I've been stuck on this for the second day now. Any help is appreciated.

peterSO · Accepted Answer · 2011-03-21 05:32:04Z

28

Here's my analysis.

(**myint)(&k) -- cannot convert &k (type **int) to type **myint:

type **int and type **myint are unnamed pointer types and their pointer base types, type *int and type *myint, don't have identical underlying types.

If T (*int or *myint) is a pointer type literal, the corresponding underlying type is T itself.

(*myint)(k) -- can convert k (type *int) to type *myint:

type *int and type *myint are unnamed pointer types and their pointer base types, type int and type myint (type myint int), have identical underlying types.

If T (int) is a predeclared type, the corresponding underlying type is T itself. If T (myint) is neither a predeclared type or nor a type literal, T's underlying type is the underlying type of the type to which T refers in its type declaration (type myint int).

(myint)(*k) -- can convert *k (type int) to type myint:

type int and type myint have identical underlying types.

If T (int) is a predeclared type, the corresponding underlying type is T itself. If T (myint) is neither a predeclared type or nor a type literal, T's underlying type is the underlying type of the type to which T refers in its type declaration (type myint int).

Here's the underlying type example from the Types section revised to use integers and int pointers.

type T1 int
type T2 T1
type T3 *T1
type T4 T3

The underlying type of int, T1, and T2 is int. The underlying type of *T1, T3, and T4 is *T1.

References:

The Go Programming Language Specification

Conversions

Types

Properties of types and values

Type declarations

Predeclared identifiers

Pointer Type

edited Mar 21, 2011 at 5:32

answered Mar 21, 2011 at 0:12

peterSO

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3 Comments

Gunchars Over a year ago

Thank you. Third line is the answer that I was looking for.

Daniel Sperry Over a year ago

I think you could add also this link: golang.org/ref/spec#Type_identity Two named types are identical if their type names originate in the same TypeSpec. so myint != int and *myint != *int

ceving Over a year ago

@DanielSperry Your statement about the pointers seems to be wrong, because the answer with the examples shows how to cast from *int to *myint and the other way round from *myint to *int. It is only necessary to be a bit more explicit in double references.

ceving · Accepted Answer · 2014-02-04 12:00:34Z

15

Here are two functionally equivalent working versions of your program.

package main

type mypint *int

func set(a *mypint) {
    i := int(5)
    *a = &i
}

func main() {
    var k *int
    set((*mypint)(&k))
    print(*k)
}

http://play.golang.org/p/l_b9LBElie

package main

type myint int

func set(a *myint) *myint {
    i := myint(5)
    a = &i
    return a
}

func main() {
    var k *int
    k = (*int)(set((*myint)(k)))
    print(*k)
}

http://play.golang.org/p/hyaPFUNlp8

edited Feb 4, 2014 at 12:00

ceving

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answered Mar 21, 2011 at 1:33

peterSO

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