When defer func evaluates its parameters

Ok I just figured out. If you pass any parameters to the defer func (like the 2nd defer function above), those parameters are evaluated when the defer function is deferred, not when they are executed. That means in my example err is still nil and has not been assigned to a new error yet.

on the other hand, in 1st defer above, err is not a parameter, but a variable in function a, and when 1st defer is executed, it has already been assigned to a new error.

There is another similar situation there is in case of Defer Statement and Defer Function. Please have look at the example below

package main

import (
    "fmt"
    "time"
)

func main() {

    start := time.Now()
    time.Sleep(3*time.Second)
    defer func() { fmt.Println("Defer Function Elapsed Time: ", time.Since(start)) }() //Defer Function
    defer fmt.Println("Defer Statement Elapsed Time: ", time.Since(start)) //Defer Statement
    time.Sleep(3*time.Second)
}

Output:

Defer Statement Elapsed Time: 3s

Defer Function Elapsed Time: 6s

Try above in go play

This is because of in the Deferred Statement case the deferred call's arguments are evaluated immediately refer doc

Another way is by using reference to original err variable

package main

import (
    "errors"
    "fmt"
)

func main() {
    a()
}

func a() {
    var err error
    defer func() {
        if err != nil {
            fmt.Printf("1st defer: %s\n", err)
        } else {
            fmt.Println("1st defer: defer not error")
        }
    }()
    defer func(err *error) {
        if *err != nil {
            fmt.Printf("2nd defer: %s\n", *err)
        } else {
            fmt.Println("2nd defer: defer not error")
        }
    }(&err)

    err = errors.New("new error")
    if err != nil {
        return
    }
}

And output is:

2nd defer: new error
1st defer: new error