Why Can't I Compare Go Structs with `>=` When Their Fields Are Comparable?

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Go Struct Comparison: Unexpected Error

The Go Programming Language Specification states that structs with only comparable fields should be comparable. However, the following code fails to compile:

type Student struct {
  Name  string // "String values are comparable and ordered, lexically byte-wise."
  Score uint8  // "Integer values are comparable and ordered, in the usual way."
}

func main() {
  alice := Student{"Alice", 98}
  carol := Student{"Carol", 72}

  if alice >= carol {
    println("Alice >= Carol")
  } else {
    println("Alice < Carol")
  }
}

The error message is:

invalid operation: alice >= carol (operator >= not defined on struct)

This error contradicts the specification, as structs should be comparable if their fields are.

Explanation:

While the fields of the Student struct can be compared (using == and !=), they are not ordered. Ordering operators (<, <=, >, >=) can only be used on operands that are ordered, such as integers or strings.

The Go Programming Language Specification clearly states that structs are comparable but not ordered:

The equality operators == and != apply to operands that are comparable. The ordering operators <, <=, >, and >= apply to operands that are ordered.

...

• Struct values are comparable if all their fields are comparable. Two struct values are equal if their corresponding non-blank fields are equal.

Therefore, you cannot use >= to compare Student structs, even though their fields are comparable.

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