CODEWITHSUNDEEP
go
Alex.Li
Alex.Li

Reputation: 560

interface & integer comparison in golang

I don't understand why the first result is false while second is true.

Any help will be appreciated.

func main() {
    var i interface{}

    i = uint64(0)
    fmt.Println("[1] ", reflect.TypeOf(i), i == 0)

    i = 0
    fmt.Println("[2] ", reflect.TypeOf(i), i == 0)

    var n uint64 = 32
    fmt.Println("[3] ", reflect.TypeOf(n), n == 32) 
}

// result
// [1]  uint64 false
// [2]  int true
// [3]  uint64 true

Try it here Go playground

Upvotes: 11

Views: 10583

Answers (1)

dave
dave

Reputation: 64657

Because 0 is an untyped constant whose default type is int, not uint64, and when doing comparison with an interface, the thing you are comparing to must be both the same type and the same value for them to be considered equal.

https://golang.org/ref/spec#Comparison_operators

The equality operators == and != apply to operands that are comparable. The ordering operators <, <=, >, and >= apply to operands that are ordered. These terms and the result of the comparisons are defined as follows:

A value x of non-interface type X and a value t of interface type T are comparable when values of type X are comparable and X implements T. They are equal if t's dynamic type is identical to X and t's dynamic value is equal to x.

Upvotes: 12

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