go - How does defer and named return value work?

Question

I just started learning Go and I got confused with one example about using defer to change named return value in the The Go Blog - Defer, Panic, and Recover.

The example says:

3. Deferred functions may read and assign to the returning function's named return values.

In this example, a deferred function increments the return value i after the surrounding function returns. Thus, this function returns 2:

func c() (i int) {
    defer func() { i++ }()
    return 1
}

But as what I have learned from A Tour of Go - Named return values

A return statement without arguments returns the named return values. This is known as a "naked" return.

I tested in the following code and in function b it returns 1 because it wasn't the "A return statement without arguments" case mentioned above.

func a() (i int) { // return 2
    i = 2
    return
}

func b() (i int) {  // return 1 
    i = 2
    return 1
}

So my question is in the first example, the surrounding function c has a named return value i, but the function c uses return 1 which in the second example we can see it should have return 1 no matter what value i is. But why after i changes in the deferred function the c function returns the value of i instead the value 1?

As I was typing my question, I might have guessed the answer. Is it because:

return 1 

is equals to:

i = 1
return 

in a function with a named return value variable i?

Please help me confirm, thanks!

Answer 1

Waitting for answers