Why Does Comparing a Nil Struct Instance to Nil Return \"Error is not nil\"?

Interface Comparisons with Nil Values

Question:

Despite declaring a nil instance of a struct that satisfies the error interface, why does the comparison to nil return "Error is not nil"?

Answer:

Interface comparisons in Go consider both the type and value of the compared entities. When comparing an error instance to nil, the following occurs:

• The nil instance of the error interface has a nil value and nil type (error(nil)).
• The nil instance of the Goof struct has a nil value but a non-nil type (*Goof).

Therefore, despite both instances being initialized with nil values, their differing types result in an unequal comparison.

To address this and return the expected "Error is nil" output, the nil instance should be assigned to an error variable, as its zero value is error(nil). Alternatively, returning nil in a function that returns an error would also produce the intended result.

Elaboration on Interface Comparisons:

Go interfaces are implemented using a type and a value. The type indicates the interface's concrete implementation, while the value represents the actual data stored. An interface value is considered nil only when both the type and value are unset (nil, nil).

When comparing two interface values for equality, both the types and values must match. In the provided example, the nil instance of the Goof struct has a nil value but a non-nil type, rendering it non-equal to the error instance, which has both a nil value and type.

This concept extends beyond nil instances. For instance, two integer values with the same underlying data (e.g., 3) but different types (int and Bob) will compare as unequal when stored in interfaces because the types differ.

The above is the detailed content of Why Does Comparing a Nil Struct Instance to Nil Return \"Error is not nil\"?. For more information, please follow other related articles on the PHP Chinese website!