Pointer Dereferencing in Go: What's Happening?
When delving into the world of programming in Go, you may come across an intriguing behavior related to pointer dereferencing, particularly when working with structs. Consider the following example:
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() { t := *q q.X = 4 u := *q fmt.Println(p, q, r, s, t, u, t == u) }
Running this code produces surprising results:
{1 2} &{4 2} {1 0} {0 0} {1 2} {4 2} false
You might expect t to change to {4, 2} after altering q.X, but this does not happen. What's the reason for this behavior?
Pointer Dereferencing Explained
The key to understanding this is pointer dereferencing. In Go, when you dereference a pointer (e.g., *q), you create a copy of the value it points to. So, t := *q makes a separate copy of the Vertex struct referenced by q.
Example 28's Modification Clarified:
In your modified example, when you set q.X = 4, you are only modifying the struct pointed to by q. Since t is a copy, it retains its original values.
Comparing Go and C/C
You mentioned the behavior seeming strange from a C/C perspective. However, C/C behave similarly. Consider this example:
#include <iostream> struct Vertex { int x; int y; }; int main() { Vertex v = Vertex{1, 2}; Vertex* q = &v; Vertex t = *q; q->x = 4; std::cout <p>This C code produces the same behavior:</p> <pre class="brush:php;toolbar:false">*q: { 4, 2 } t: { 1, 2 }
Conclusion
In summary, when you dereference a pointer in Go, you're creating a copy of the underlying value. To observe changes made through a pointer, you must use a pointer itself, like in the modified example:
func main() { t := q q.X = 4 u := *q fmt.Println(p, q, r, s, t, u, *t == u) }
This will produce the expected output of {1 2} &{4 2} {1 0} {0 0} {4 2} {4 2} true.
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