Pointers in Go

In Go, an unnamed pointer type can be represented with *T, where T can be an arbitrary type. Type T is called the base type of pointer type *T.

We can declare named pointer types, but generally, it’s not recommended to use named pointer types, for unnamed pointer types have better readabilities.

If the underlying type of a named pointer type is *T, then the base type of the named pointer type is T.

Two unnamed pointer types with the same base type are the same type.

Example:

*int  // An unnamed pointer type whose base type is int.
**int // An unnamed pointer type whose base type is *int.

// Ptr is a named pointer type whose base type is int.
type Ptr *int
// PP is a named pointer type whose base type is Ptr.
type PP *Ptr

Zero values of any pointer types are represented with the predeclared nil. No addresses are stored in nil pointer values.

A value of a pointer type whose base type is T can only store the addresses of values of type T.

There are two ways to get a non-nil pointer value.

1. The built-in new function can be used to allocate memory for a value of any type. new(T) will allocate memory for a T value (an anonymous variable) and return the address of the T value. The allocated value is a zero value of type T. The returned address is viewed as a pointer value of type *T.
2. We can also take the addresses of values which are addressable in Go. For an addressable value t of type T, we can use the expression &t to take the address of t, where & is the operator to take value addresses. The type of &t is viewed as *T.

The following program shows some address taking and pointer dereference examples:

package main

import "fmt"

func main() {
	p0 := new(int)   // p0 points to a zero int value.
	fmt.Println(p0)  // (a hex address string)
	fmt.Println(*p0) // 0

	// x is a copy of the value at
	// the address stored in p0.
	x := *p0
	// Both take the address of x.
	// x, *p1 and *p2 represent the same value.
	p1, p2 := &x, &x
	fmt.Println(p1 == p2) // true
	fmt.Println(p0 == p1) // false
	p3 := &*p0 // <=> p3 := &(*p0) <=> p3 := p0
	// Now, p3 and p0 store the same address.
	fmt.Println(p0 == p3) // true
	*p0, *p1 = 123, 789
	fmt.Println(*p2, x, *p3) // 789 789 123

	fmt.Printf("%T, %T \n", *p0, x) // int, int
	fmt.Printf("%T, %T \n", p0, p1) // *int, *int
}

The following picture depicts the relations of the values used in the above program.

Let's view an example firstly.

package main

import "fmt"

func double(x int) {
	x += x
}

func main() {
	var a = 3
	double(a)
	fmt.Println(a) // 3
}

The double function in the above example is expected to modify the input argument by doubling it. However, it fails. Why? Because all value assignments, including function argument passing, are value copying in Go. What the double function modified is a copy (x) of variable a but not variable a.

One solution to fix the above double function is let it return the modification result. This solution doesn't always work for all scenarios. The following example shows another solution, by using a pointer parameter.

package main

import "fmt"

func double(x *int) {
	*x += *x
	x = nil // the line is just for explanation purpose
}

func main() {
	var a = 3
	double(&a)
	fmt.Println(a) // 6
	p := &a
	double(p)
	fmt.Println(a, p == nil) // 12 false
}

We can find that, by changing the parameter to a pointer type, the passed pointer argument &a and its copy x used in the function body both reference the same value, so the modification on *x is equivalent to a modification on *p, a.k.a., variable a. In other words, the modification in the double function body can be reflected out of the function now.

Surely, the modification of the copy of the passed pointer argument itself still can't be reflected on the passed pointer argument. After the second double function call, the local pointer p doesn't get modified to nil.

In short, pointers provide indirect ways to access some values. Many languages do not have the concept of pointers. However, pointers are just hidden under other concepts in those languages.

Unlike C language, Go is a language supporting garbage collection, so return the address of a local variable is absolutely safe in Go.

func newInt() *int {
	a := 3
	return &a
}

In Go, pointers can't do arithmetic operations. For a pointer p, p++ and p-2 are both illegal.

If p is a pointer to a numeric value, compilers will view *p++ is a legal statement and treat it as (*p)++. In other words, the precedence of the pointer dereference operator * is higher than the increment operator ++ and decrement operator --.

Example:

package main

import "fmt"

func main() {
	a := int64(5)
	p := &a

	// The following two lines don't compile.
	/*
	p++
	p = (&a) + 8
	*/

	*p++
	fmt.Println(*p, a)   // 6 6
	fmt.Println(p == &a) // true

	*&a++
	*&*&a++
	**&p++
	*&*p++
	fmt.Println(*p, a) // 10 10
}

In Go, a pointer value of pointer type T1 can be directly and explicitly converted to another pointer type T2 only if either of the following two conditions is get satisfied.

1. The underlying types of type T1 and T2 are identical (ignoring struct tags), in particular if either T1 and T2 is a unnamed type and their underlying types are identical (considering struct tags), then the conversion can be implicit. Struct types and values will be explained in the next article.
2. Type T1 and T2 are both unnamed pointer types and the underlying types of their base types are identical (ignoring struct tags).

For example, for the below shown pointer types:

type MyInt int64
type Ta    *int64
type Tb    *MyInt

the following facts exist:

1. values of type *int64 can be implicitly converted to type Ta, and vice versa, for their underlying types are both *int64.
2. values of type *MyInt can be implicitly converted to type Tb, and vice versa, for their underlying types are both *MyInt.
3. values of type *MyInt can be explicitly converted to type *int64, and vice versa, for they are both unnamed and the underlying types of their base types are both int64.
4. values of type Ta can't be directly converted to type Tb, even if explicitly. However, by the just listed first three facts, a value pa of type Ta can be indirectly converted to type Tb by nesting three explicit conversions, Tb((*MyInt)((*int64)(pa))).

None values of these pointer types can be converted to type *uint64, in any safe ways.

In Go, pointers can be compared with == and != operators. Two Go pointer values can only be compared if either of the following three conditions are satisfied.

1. The types of the two Go pointers are identical.
2. One pointer value can be implicitly converted to the pointer type of the other. In other words, the underlying types of the two types must be identical and either of the two types of the two Go pointers is an unnamed type.
3. One and only one of the two pointers is represented with the bare (untyped) nil identifier.

Example:

package main

func main() {
	type MyInt int64
	type Ta    *int64
	type Tb    *MyInt

	// 4 nil pointers of different types.
	var pa0 Ta
	var pa1 *int64
	var pb0 Tb
	var pb1 *MyInt

	// The following 6 lines all compile okay.
	// The comparison results are all true.
	_ = pa0 == pa1
	_ = pb0 == pb1
	_ = pa0 == nil
	_ = pa1 == nil
	_ = pb0 == nil
	_ = pb1 == nil

	// None of the following 3 lines compile ok.
	/*
	_ = pa0 == pb0
	_ = pa1 == pb1
	_ = pa0 == Tb(nil)
	*/
}

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