pass-a-struct-to-a-function.md

Pass A Struct To A Function

Go operates as pass-by-value which means that when we pass a struct to a function, the receiving function gets a copy of the struct. Two things worth noticing about that are 1) an extra memory allocation happens when calling the function and 2) altering the struct does not affect the original in the calling context.

On the other hand, we can have a function that takes a pointer to a struct. When we call that function, we have a reference to the memory location of the struct instead of a copy of the struct. That means no additional allocation and modifications to the dereferenced struct are modifications to the original in the calling context.

Here is an example that demonstrates both of these. Notice the printed output that is included in comments at the end which shows memory locations and contents of the struct at various points.

package main

import "fmt"

type Order struct {
	Item string
	Quantity int
	DineIn bool
}

func main() {
	order := Order{Item: "taco", Quantity: 3, DineIn: true}

	fmt.Println("Order:", order)
	fmt.Printf("main - Loc: %p\n", &order)

	doubledOrder := doubleOrder(order)

	fmt.Println("Double Order:", doubledOrder)
	fmt.Println("Original Order:", order)

	doubleOrderPtr(&order)

	fmt.Println("Double Order Ptr:", order)
}

func doubleOrder(order Order) Order {
	fmt.Printf("doubleOrder - Loc: %p\n", &order)
	order.Quantity *= 2

	return order
}

func doubleOrderPtr(order *Order) {
	fmt.Printf("doubleOrderPtr - Loc: %p\n", order)
	(*order).Quantity *= 2
}

// Order: {taco 3 true}
// main - Loc: 0xc0000b4000
// doubleOrder - Loc: 0xc0000b4040
// Double Order: {taco 6 true}
// Original Order: {taco 3 true}
// doubleOrderPtr - Loc: 0xc0000b4000
// Double Order Ptr: {taco 6 true}