What does this print, and why? errors.New("x") == errors.New("x")

a := errors.New("x")
b := errors.New("x")
fmt.Println(a == b)             // false
fmt.Println(a.Error() == b.Error()) // true

type codeErr struct{ code int }
func (e codeErr) Error() string { return strconv.Itoa(e.code) }

var c, d error = codeErr{1}, codeErr{1}
fmt.Println(c == d)             // true

errors.New returns a *errorString, a pointer. Two interface values are equal when their dynamic types and dynamic values are equal. Here the values are two different pointers, so each errors.New call creates a unique identity. That is why sentinel errors such as io.EOF work: you compare against the one package-level variable.

With a value type (codeErr), equality compares fields, so two separately created errors are ==. That may be what you want, but it can surprise you. It also means a sentinel of a struct value type can be "forged" by any caller.

Gotcha: never compare errors by their Error() string. Messages are for humans and can change.