Why does a nil instance of a struct that satisfies the error interface not compare equal to nil?

Nil Nil Interface Comparison Puzzle

Despite satisfying the error interface, a struct with a nil instance does not behave as expected when compared to nil.

Question:

Why does the following code not output "Error is nil"?

<code class="go">type Goof struct {}

func (goof *Goof) Error() string {
    return fmt.Sprintf("I'm a goof")
}

func TestError(err error) {
    if err == nil {
        fmt.Println("Error is nil")
    } else {
        fmt.Println("Error is not nil")
    }
}

func main() {
    var g *Goof // nil
    TestError(g) // expect "Error is not nil"
}</code>

Answer:

In Go, interface comparisons consider both type and value. While the Goof type implements the error interface, a nil instance of Goof (*Goof)(nil) has a distinct type from error(nil).

Solution:

To address this, you can adopt one of the following approaches:

• Declare err error instead of var g *Goof. This initializes err to the zero value of the error type, which is nil.
• In functions returning errors, return nil explicitly instead of leaving it implicit.

For more details, refer to the extended response below:

Extended Response:

Interface values consist of two components: a type and a dynamic value. A nil interface value contains both a nil type and a nil value. In our case, (*Goof)(nil) has a non-nil type (Goof), but a nil value.

Additionally, Go's equality operator (==) strictly checks for type identity. Hence, comparing (*Goof)(nil) to error(nil) fails because they are of different types.

This behavior is consistent with other type-checking in Go. For instance, in the following code, the underlying data is the same (3), but the variables have different types, resulting in inequality when stored in interfaces:

<code class="go">var x int = 3
var y Bob = 3
var ix, iy interface{} = x, y
fmt.Println(ix == iy) // false</code>

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