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How Can Deferred Functions Modify Named Return Values in Go?

Mary-Kate Olsen
Mary-Kate OlsenOriginal
2024-12-03 08:36:10341browse

How Can Deferred Functions Modify Named Return Values in Go?

Understanding the Interplay of Defer and Named Return Values

In Go, the defer statement serves as a powerful tool for postponing function calls until just before the enclosing function exits. This capability can be leveraged to modify the named return values of a function, as illustrated in the Go Blog example:

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

In this example, the defer statement schedules a function that increments the named return value i after the surrounding function c returns. The question arises: how can a deferred function alter the return value of the enclosing function?

The answer lies in Go's handling of return statements. In functions with named return values, a return statement without explicit arguments assigns the current values of those named variables to the return values. In the case of function c, the return statement return 1 is equivalent to i = 1; return.

Prior to the introduction of the defer statement, the return statement effectively assigns the value 1 to the named variable i and then immediately returns from the function. However, with the defer statement in place, the deferred function is called before the surrounding function returns. This deferred function increments i, thus altering its value. As a result, the return statement ultimately returns the modified value of i, which is 2.

This behavior demonstrates the power of deferred functions to not only modify variable values after a function has returned but also to alter the named return values of the enclosing function.

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