How do I print the pointer value of a Go object? What does the pointer value mean?

Solution 1:

Go function arguments are passed by value.

First, let's discard the irrelevant parts of your example, so that we can easily see that you are merely passing an argument by value. For example,

package main

import "fmt"

func byval(q *int) {
    fmt.Printf("3. byval -- q %T: &q=%p q=&i=%p  *q=i=%v\n", q, &q, q, *q)
    *q = 4143
    fmt.Printf("4. byval -- q %T: &q=%p q=&i=%p  *q=i=%v\n", q, &q, q, *q)
    q = nil
}

func main() {
    i := int(42)
    fmt.Printf("1. main  -- i  %T: &i=%p i=%v\n", i, &i, i)
    p := &i
    fmt.Printf("2. main  -- p %T: &p=%p p=&i=%p  *p=i=%v\n", p, &p, p, *p)
    byval(p)
    fmt.Printf("5. main  -- p %T: &p=%p p=&i=%p  *p=i=%v\n", p, &p, p, *p)
    fmt.Printf("6. main  -- i  %T: &i=%p i=%v\n", i, &i, i)
}

Output:

1. main  -- i  int: &i=0xf840000040 i=42
2. main  -- p *int: &p=0xf8400000f0 p=&i=0xf840000040  *p=i=42
3. byval -- q *int: &q=0xf8400000d8 q=&i=0xf840000040  *q=i=42
4. byval -- q *int: &q=0xf8400000d8 q=&i=0xf840000040  *q=i=4143
5. main  -- p *int: &p=0xf8400000f0 p=&i=0xf840000040  *p=i=4143
6. main  -- i  int: &i=0xf840000040 i=4143

In function main, i is an int variable at memory location (&i) 0xf800000040 with an initial value (i) 42.

In function main, p is a pointer to an int variable at memory location (&p) 0xf8000000f0 with a value (p=&i) 0xf800000040 which points to an int value (*p=i) 42.

In function main, byval(p) is a function call which assigns the value (p=&i) 0xf800000040 of the argument at memory location (&p) 0xf8000000f0 to the function byval parameter q at memory location (&q) 0xf8000000d8. In other words, memory is allocated for the byval parameter q and the value of the main byval argument p is assigned to it; the values of p and q are initially the same, but the variables p and q are distinct.

In function byval, using pointer q (*int), which is a copy of pointer p (*int), integer *q (i) is set to a new int value 4143. At the end before returning. the pointer q is set to nil (zero value), which has no effect on p since q is a copy.

In function main, p is a pointer to an int variable at memory location (&p) 0xf8000000f0 with a value (p=&i) 0xf800000040 which points to a new int value (*p=i) 4143.

In function main, i is an int variable at memory location (&i) 0xf800000040 with a final value (i) 4143.

In your example, the function main variable s used as an argument to the function gotest call is not the same as the function gotest parameter s. They have the same name, but are different variables with different scopes and memory locations. The function parameter s hides the function call argument s. That's why in my example, I named the argument and parameter variables p and q respectively to emphasize the difference.

In your example, (&s) 0x4930d4 is the address of the memory location for the variable s in function main that is used as an argument to the function call gotest(s, done), and 0x4974d8 is the address of the memory location for the function gotest parameter s. If you set parameter s = nil at the end of function gotest, it has no effect on variable s in main; s in main and s in gotest are distinct memory locations. In terms of types, &s is **Something, s is *Something, and *s is Something. &s is a pointer to (address of memory location) s, which is a pointer to (address of memory location) an anonymous variable of type Something. In terms of values, main.&s != gotest.&s, main.s == gotest.s, main.*s == gotest.*s, and main.s.number == gotest.s.number.

You should take mkb's sage advice and stop using println(&s). Use the fmt package, for example,

fmt.Printf("%v %p %v\n", &s, s, *s)

Pointers have the same value when they point to the same memory location; pointers have different values when they point to different memory locations.

Solution 2:

In Go, arguments are passed by value.

package main

import "fmt"

type SomeStruct struct {
    e int
}

// struct passed by value
func v(v SomeStruct) {
    fmt.Printf("v: %p %v\n", &v, v)
    v.e = 2
    fmt.Printf("v: %p %v\n", &v, v)
}

// pointer to struct passed by value
func p(p *SomeStruct) {
    fmt.Printf("p: %p %v\n", p, *p)
    p.e = 2
    fmt.Printf("p: %p %v\n", p, *p)
}

func main() {
    var s SomeStruct
    s.e = 1
    fmt.Printf("s: %p %v\n", &s, s)
    v(s)
    fmt.Printf("s: %p %v\n", &s, s)
    p(&s)
    fmt.Printf("s: %p %v\n", &s, s)
}

Output:

s: 0xf800000040 {1}
v: 0xf8000000e0 {1}
v: 0xf8000000e0 {2}
s: 0xf800000040 {1}
p: 0xf800000040 {1}
p: 0xf800000040 {2}
s: 0xf800000040 {2}

Solution 3:

type sometype struct { }
a := sometype {}
b := int(2)
println("Ptr to a", &a)
println("Ptr to b", &b)

Solution 4:

How do I print the pointer value of a Go object?

package main

import (
    "fmt"
)

func main() {
    a := 42
    fmt.Println(&a)
}

results in:

0x1040a124

What does the pointer value mean?

According to Wikipedia:

A pointer references a location in memory

Solution 5:

package main

import "fmt"

func zeroval(ival int) {
     ival = 0
}

func zeroptr(iptr *int) {
     *iptr = 0
}

func main() {
    i := 1
    fmt.Println("initial:", i)
    zeroval(i)
    fmt.Println("zeroval:", i)
    //The &i syntax gives the memory address of i, i.e. a pointer to i.
    zeroptr(&i)
    fmt.Println("zeroptr:", i)
    //Pointers can be printed too.
    fmt.Println("pointer:", &i)
}

OUTPUT:

$ go run pointers.go
initial: 1
zeroval: 1
zeroptr: 0
pointer: 0x42131100