How Do `defer` Statements Affect Named Return Values in Go?

How does defer interact with named return values?

In Go, the defer keyword defers the execution of a function until the surrounding function returns. In the context of named return values, this behavior allows deferred functions to modify the values that will ultimately be returned.

Consider the following example:

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

In this example, the main function returns 1. However, the deferred function increments the named return value i by 1 before the main function returns. As a result, the actual return value of the main function is 2.

This behavior is possible because in Go, return statements with arguments effectively assign values to named return variables before returning. Therefore, the statement return 1 is equivalent to the following:

i = 1
return

As a result, the deferred function can assign to the named return value i and modify the return value of the surrounding function.

It's important to note that the ordering of defer statements is important. Deferred functions are executed in a Last-In-First-Out (LIFO) order. This means that the most recently deferred function will be executed first after the surrounding function returns.

To better understand how this works, consider the following revised example:

func c() (i int) {
    defer func() { fmt.Println("third") }()
    defer func() { i = 2 }()
    defer func() { fmt.Println("first") }()

    fmt.Println("second")
    return 1
}

In this example, the order of execution is as follows:

1. fmt.Println("second")
2. return 1
3. i = 2
4. fmt.Println("first")
5. fmt.Println("third")

The deferred function that sets i to 2 is executed before the function returns, so the return value is 2. The fmt.Println statements are executed in reverse order due to the LIFO behavior of defer.

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