Does Dereferencing a Go Struct Create a New Copy?

Does Dereferencing a Struct Create a New Copy in Go?

In Go, when a variable of a struct type is passed by reference using the asterisk (*) operator, a misconception may arise that the dereferenced variable represents the same address as the original struct.

However, as demonstrated in the given code snippet:

package main

import (
    "fmt"
)

type me struct {
    color string
    total int
}

func study() *me {
    p := me{}
    p.color = "tomato"
    fmt.Printf("%p\n", &p.color) // Address of p.color
    return &p
}

func main() {
    p := study() // p is a pointer to a struct
    fmt.Printf("&p.color = %p\n", &p.color) // Address of p.color

    obj := *p // obj is a copy of the struct pointed to by p
    fmt.Printf("&obj.color = %p\n", &obj.color) // Address of obj.color

    fmt.Printf("obj = %+v\n", obj)

    p.color = "purple" // Changes color of the struct pointed by p
    fmt.Printf("p.color = %p\n", &p.color)
    fmt.Printf("p = %+v\n", p)
    fmt.Printf("obj  = %+v\n", obj)

    obj2 := *p // Another copy is made
    fmt.Printf("obj2 = %+v\n", obj2)
}

When we execute this code, the output shows that the dereferenced variable obj has a different address than the original struct p. This is because:

Dereferencing creates a new copy:

The line obj := *p creates a new variable obj of the same type as p (me), and initializes it with a copy of the value pointed by p. This means that any changes made to obj will not affect the original struct.

Effectively assigning a new struct value to another:

Similar to var obj me = *p, the dereference operation in obj := *p assigns a new struct value to the variable obj. This creates a new copy with separate memory.

Using the asterisk operator with caution:

While the asterisk operator provides pointer semantics, it's important to understand its effects on variables. When a value is assigned with the * (dereference) operator, it creates a new copy, not a new reference to the original variable.

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