Why Does Struct Comparison Fail in Go Despite Comparable Fields?

Go Struct Comparison: Why the Expected Comparison Fails

Despite conforming to the Go specification on struct comparability, where all fields of a struct are comparable, the following code fails to compile:

type Student struct {
  Name  string 
  Score uint8  
}

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

  if alice >= carol { // Error: "invalid operation"
    println("Alice >= Carol")
  } else {
    println("Alice <h2 id="The-Missing-Distinction-Comparable-vs-Ordered">The Missing Distinction: Comparable vs. Ordered</h2><p>The key to understanding this failure lies in distinguishing between comparability and orderability.</p>

• Comparable: structs are comparable if all their fields are comparable.
• Ordered: structs are ordered if all their fields are ordered, allowing for operators like , and >= to be used.

While the Student struct's fields are indeed comparable, the >= operator is an ordering operator, not a comparable one. This distinction is explicitly mentioned in the Go specification, where struct values are defined as "comparable if all their fields are comparable," but only "equal if their corresponding non-blank fields are equal." Ordering is not mentioned in this context.

Conclusion

The misconception that led to this compile-time error stems from assuming that comparability implies orderability. In Go, these are distinct concepts. While structs can be compared for equality, they cannot be ordered by default unless their fields provide a specific ordering relation.

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