CODEWITHSUNDEEP
pointersgostructreferencepass-by-reference
21st
21st

Reputation: 2111

Why does the memory address of a struct reference change?

I have a struct and a method that's working on the structs reference. The pointer address is changing every time I call the method. Why is it like that?

Code

package main

import "k8s.io/contrib/compare/Godeps/_workspace/src/github.com/emicklei/go-restful/log"

type Whatever struct{
    Name string
}

func (whatever *Whatever) GetNameByReference() (string) {
log.Printf("Whatever.GetNameByReference() memory address: %v", &whatever)

    return whatever.Name
}

func evaluateMemoryAddressWhenNotWritingAnything()  {
    whatever := Whatever{}

    whatever.GetNameByReference()
    whatever.GetNameByReference()
    whatever.GetNameByReference()
}

func main() {
    evaluateMemoryAddressWhenNotWritingAnything()
}

Output:

log.go:30: Whatever.GetNameByReference() memory address: 0xc420034020
log.go:30: Whatever.GetNameByReference() memory address: 0xc420034030
log.go:30: Whatever.GetNameByReference() memory address: 0xc420034038

Upvotes: 5

Views: 3794

Answers (3)

Uvelichitel
Uvelichitel

Reputation: 8490

Calling method

func (whatever *Whatever) GetNameByReference() (string) {

is much the same as calling function providing receiver as first argument

func GetNameByReference(whatever *Whatever) (string) {

Go calling convention is to always pass arguments copy_by_value. So each time you call whatever.GetNameByReference() it receives fresh copy of whatever which itself holds same *Whatever address. As already written doing

log.Printf("Whatever.GetNameByReference() memory address: %v", whatever) //instead of &whatever

will log the same whatever value which is address.

Upvotes: 0

Volker
Volker

Reputation: 42413

Never think and never talk about references. Go has no notion of "reference", everything is a value. Some things are pointer values. Your problem stems from thinking about *X as "a reference to an X" which it isn't: It is a value holding the memory address of an X (or nil).

So in func (whatever *Whatever) the variable whatever is a pointer to a Whatever. The value of whatever is the memory address of the Whatever the pointer points to. You would like to print this memory address, i.e. the value of whatever.

You do Printf("%v", &whatever). Remember: whatever is a variable (holding a memory address). So &whatever is the address of the variable itself: &whatever is of type **Whatever. What you find at the address &whatever is not the value you are interested in; it is just the temporary variable used to store the address of the original Whatever. Of course this temporary variable is not pinned in memory an may change freely.

You should do Printf("%p", whatever). The verb %p is for pointer values and whateveris a pointer and you are interested in its value, so print this value.

Upvotes: 12

Franck Jeannin
Franck Jeannin

Reputation: 6844

You are not displaying the address of the struct, you are displaying the address of the address (address of the pointer). The pointer is passed as a parameter and thus new each time. Remove the & in log.Printf("Whatever.GetNameByReference() memory address: %v", &whatever) to get what you want (and use %p instead of %v).

Upvotes: 3

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