How does pointer dereferencing work in Go?

Question

I'm going through the golang tutorials at http://tour.golang.org/, and was experimenting a bit with some things in example 29

For your reference, the original example is copied here:

package main  import "fmt"  type Vertex struct {     X, Y int }  var (     p = Vertex{1, 2}  // has type Vertex     q = &Vertex{1, 2} // has type *Vertex     r = Vertex{X: 1}  // Y:0 is implicit     s = Vertex{}      // X:0 and Y:0 )  func main() {     fmt.Println(p, q, r, s) } 

It's pretty basic, showing how to create instances of this fancy new struct, Vertex. Example 28, though, shows manipulating a vertex via a pointer to it, so I modified the example a bit and was surprised by the output. Here is the modification:

func main() {     t := *q     q.X = 4     u := *q     fmt.Println(p, q, r, s, t, u, t == u) } 

And the output:

{1 2} &{4 2} {1 0} {0 0} {1 2} {4 2} false 

The thing that surprised me is that t is not {4, 2}, which seems to mean that changing q.X changed the instance of the struct that q pointed to. Coming from a C/C++ background, this seems like extremely strange behavior to me.

So, what's actually going on here? Why does using q.X = 4 to change the Vertex not propagate to t?

Answer 1

t := *q makes a copy of the struct pointed to by q.

If you want to observe changes to q through t, then stick with a pointer:

var (     p = Vertex{1, 2}  // has type Vertex     q = &Vertex{1, 2} // has type *Vertex     r = Vertex{X: 1}  // Y:0 is implicit     s = Vertex{}      // X:0 and Y:0 )   func main() {     t := q     q.X = 4     u := *q     fmt.Println(p, q, r, s, t, u, *t == u) } 

This produces the output you were probably looking for.

{1 2} &{4 2} {1 0} {0 0} &{4 2} {4 2} true 

I'm not sure what seems extremely strange to you. C and C++ behave the same way. Consider the following:

#include <iostream>  struct Vertex {     int x;     int y; };  std::ostream& operator<<(std::ostream& out, const Vertex& v) {     out << "{ " << v.x << ", " << v.y << " }";      return out; }  int main() {     Vertex v = Vertex{1, 2};     Vertex* q = &v;     Vertex t = *q;     q->x = 4;     std::cout << "*q: " << *q << "\n";     std::cout << " t: " << t << "\n"; } 

The output of this C++ code shows the same behavior:

*q: { 4, 2 }   t: { 1, 2 }