为什么Go的结果相同？

Question

When i review a Golang project, i found some code like this, now is Unix timestap. Call this function once every second:

//main.go
func PrevSlot(now int64) int64 {
    // now = time.Now().Unix()
    var blockInterval = int64(10)

    result := int64((now-1)/blockInterval) * blockInterval // why result is same ?
    plog.Println("PrevSlot:", int64(result), int64((now-1)/blockInterval), now)
    return result
}


func main() {
    ticker := time.NewTicker(1 * time.Second)
    for {
        <-ticker.C

        now := time.Now().Unix()
        PrevSlot(now)
    }

}

Output:

2019-01-16 10:58:31.668597 I | dpos.go: PrevSlot: 1547607510 154760751 1547607511
2019-01-16 10:58:32.668649 I | dpos.go: PrevSlot: 1547607510 154760751 1547607512
2019-01-16 10:58:33.668568 I | dpos.go: PrevSlot: 1547607510 154760751 1547607513
2019-01-16 10:58:34.668572 I | dpos.go: PrevSlot: 1547607510 154760751 1547607514
2019-01-16 10:58:35.668547 I | dpos.go: PrevSlot: 1547607510 154760751 1547607515

The result is same. Why is this, what is the principle?

Answer 1

Actually,

int64((now - 1)/blockInterval * blockInterval

does not return the same result all the time. You will notice it changing every 10 seconds.

This is caused by integer division in Go. If you apply integer division to any two numbers, then the fractional part of the result (remainder) is dropped. For example, int(12 / 10) = 1. In your specific case - dividing by ten, the code will drop the remainders from 1 until 9, and only increment the value when you get to the next 10.

If you'd like to increase the precision of your operations, then you can force floating point division like so float64(12)/float64(10).