Why Can't I Compare Go Structs with `>=` or `

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Go Struct Comparison: Misunderstanding between Comparable and Ordered

In Go, structs are considered comparable by default if all their fields are comparable. This means a struct can be assigned to a variable of the same type or compared with another struct of the same type using the equality operator (== or !=).

However, a common misconception arises when attempting to use ordered operators such as >= or <= on structs. While structs are comparable, they are not inherently ordered. This distinction is crucial to understand for effective Go programming.

In the example provided:

package main

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 code fails to compile because >= is an ordered operator. The compiler detects that Student is not explicitly defined as an ordered type, which is necessary for it to support ordered comparisons. Struct values can be compared for equality using == or !=, but their order relationship (i.e., <, >, <=, >=) is undefined.

The Go Language Specification clearly states:

"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, structs are not intrinsically ordered, and ordered operators like >= are not supported on struct values unless they are explicitly declared as ordered types. To enable ordered comparison, custom types must implement the sort.Interface interface, which requires defining specific methods for sorting and comparing instances of the type.

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