Does a constant returned by a function automatically become the value of the named return variable

This is a function in Golang that uses defer to alter the named return value of the function c().

package main

import "fmt"

func c() (i int) { 
    defer func() {  }()     
    defer fmt.Println("our i is", i)
    return 45
   }

func main() { 	
   fmt.Println(c()) 
 }

The output of the program is:

> our i is 0
>
> 45

changing the anonymous func() in the code

func c() (i int) { 
        defer func() { i = 1 }()   
        defer fmt.Println("our i is", i)
        return 45
       }
    
    func main() { 	
       fmt.Println(c()) 
     }

This results in the output:

> our i is 0
>
> 1

It seems like the return value 45 is copied to i automatically if no other value is being put inside i. But I am not a 100% sure if that is the exact reason for the output

In deferred functions you have the opportunity to modify the values of the result parameters.

When deferred functions are called, values specified in the return statement are already set.

If there are multiple deferred functions, they are executed in LIFO order (last in, first out).

In your second example fmt.Println() is executed first, then the other, anonymous function.

But what you need to know that when the defer statement is executed, the arguments to the deferred function are evaluated immediately, not when the deferred function will be run (later, before returning).

Spec: Defer statements:

> Each time a "defer" statement executes, the function value and parameters to the call are evaluated as usual and saved anew but the actual function is not invoked. Instead, deferred functions are invoked immediately before the surrounding function returns, in the reverse order they were deferred.

So this line:

defer fmt.Println("our i is", i)

Will always mean to call fmt.Println() with i = 0 argument:

fmt.Println("our i is", 0)

Because when this line runs, i has a value of 0.

So in your 2nd example fmt.Println() prints 0, then runs the other deferred function which sets i to 1 and this is what is returned.

Your first example just prints something (i = 0), but the deferred functions in the first example do not modify the value of i so 45 will be returned (and printed by the fmt.Println() function from the main()).

Please see the following questions/answer for more on the topic:

https://stackoverflow.com/questions/31404471/defer-usage-clarification

https://stackoverflow.com/questions/33167282/how-to-return-a-value-in-a-go-function-that-panics