CODEWITHSUNDEEP
go
quietnan
quietnan

Reputation: 13

Unexpected behavior on Ticker interval change

I naively wrote a piece of code to change the interval of a time.Ticker, and upon reviewing it I am irritated that it works:

package main

import (
    "fmt"
    "time"
)

func main() {
    ticker := time.NewTicker(5 * time.Second)
    switcher := time.NewTimer(16 * time.Second)
    for {
        select {
        case <-ticker.C:
            fmt.Println("ticker " + time.Now().Format("15:04:05"))
        case <-switcher.C:
            fmt.Println("switching")
            ticker = time.NewTicker(1 * time.Second)
        }
    }
}

$ go run main.go
ticker 02:19:03
ticker 02:19:08
ticker 02:19:13
switching
ticker 02:19:15
ticker 02:19:16
ticker 02:19:17

It does what I want, namely switch the frequency of the ticker Ticker to one second after 16 seconds. However, reading it again I would expect that after 16 seconds, a new Ticker object is created, while the select statement holds a reference to the channel of the old one, preventing it from garbage collection and keep on ticking every five seconds.

So why does this code change the ticker's frequency?

Upvotes: 1

Views: 213

Answers (1)

Mr_Pink
Mr_Pink

Reputation: 109438

A select statement isn't a closure and doesn't "hold" a reference to anything. You're simply replacing the value of ticker, and as you can see, the interval changes accordingly.

Upvotes: 5

Related Questions