为什么结构引用的内存地址会更改？

Question

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

Answer 1

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.