What is the difference between named return value and normal return value?

Question

My question is about the different named return value vs normal return value.

my code

package main

import "fmt"

func main() {
    f := fmt.Println
    f(a())
    f(b())
}

func a() int {
    i := 0
    defer func() {
        i += 1
        fmt.Println("a defer : ", i)
    }()

    return i
}

func b() (i int) {
    i = 0

    defer func() {
        i += 1
        fmt.Println("b defer : ", i)
    }()
    return i
}

the result:

the a function return 0

the b function reutrn 1

Why?

Answer 1

The named return value also allocates a variable for the scope of your function.

func a() int: While you already return the value of i = 0, but since no named values was defined the static value got returned. So even though you're increasing i in the deferred function it doesn't affect the returned value.

func b() (i int): The variable i is allocated (and already initialized to 0). Even though the deferred function runs after the i = 0 was returned the scope is still accessible and thus still can be changed.

---

Another point of view: you can still change named return values in deferred functions, but cannot change regular return values.

This especially holds true in the following example:

func c() (i int) {
    defer func() {
        i = 1
        fmt.Println("c defer : ", i)
    }()
    return 0
}