How Do Go\'s `defer` Statements Affect Return Values Based on Variable Declarations?

Understanding Defer and Return Values in Go: Why Results Differ with Variable Declarations

In Go, the defer statement is used to schedule a function to be executed after the surrounding function returns. However, when it comes to accessing and modifying variable values within the deferred function, the behavior can vary depending on how the variables are declared.

Case 1: Variable Declared as Incoming Parameter

Consider the following example:

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

Here, i is declared as an incoming parameter. When the return statement is executed, the return value is evaluated, and the deferred function runs after this. As a result, incrementing i in the deferred function has no effect on the return value, and c(0) will print 0.

Case 2: Variable Declared as Result Parameter

Now let's look at this example:

func c1() (i int) {
    defer func() { i++ }()
    return i
}

In this case, i is declared as the name of the result parameter. When the return statement is executed, the value of i is explicitly returned. The deferred function is then allowed to modify the value of i, which affects the actual returned value. This is why c1() prints 1.

Case 3: Result Parameter with Explicit Return Value

To illustrate further, consider this example:

func c2() (i int) {
    defer func() { i++ }()
    return 2
}

Here, even though the defer function modifies i, the explicit return statement assigns the value 2 to i before the defer function runs. As a result, c2() returns 3.

Conclusion

The key takeaway is that in Go, if a function has named result parameters, the return values are the values of those variables. However, a return statement may assign new values to these result parameters, and deferred functions can modify them after the return statement, affecting the actual returned values.

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