doumin4553 2018-12-13 22:05
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I'm starting out with Go and I'm now writing a simple program which reads out data from a sensor and puts that into a channel to do some calculations with it. I now have it working as follows:

package main

import (

func get_sensor_data(c chan float64) {
    time.Sleep(1 * time.Second)  // wait a second before sensor data starts pooring in
    c <- 2.1  // Sensor data starts being generated
    c <- 2.2
    c <- 2.3
    c <- 2.4
    c <- 2.5

func main() {

    s := 1.1

    c := make(chan float64)
    go get_sensor_data(c)

    for {
        select {
        case s = <-c:
            fmt.Println("the next value of s from the channel: " + strconv.FormatFloat(s, 'f', 1, 64))
            // no new values in the channel

        time.Sleep(500 * time.Millisecond)  // Do heavy "work"

This works fine, but the sensor generates a lot of data, and I'm always only interested in the latest data. With this setup however, it only reads out the next item with every loop, which means that if the channel at some point contains 20 values, the newest value only is read out after 10 seconds.

Is there a way for a channel to always only contain one value at a time, so that I always only get the data I'm interested in, and no unnecessary memory is used by the channel (although the memory is the least of my worries)?

All tips are welcome!

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  • douhanzhen8927 2018-12-13 22:15

    Channels are best thought of as queues (FIFO). Therefore you can't really skip around. However there are libraries out there that do stuff like this: is an atomic ring buffer that will overwrite old data. You can set a smaller size if you like.

    All that being said, it doesn't sound like you need a queue (or ring buffer). You just need a mutex:

    type SensorData struct{
      mu sync.RWMutex
      last float64
    func (d *SensorData) Store(data float64) {
     defer mu.Unlock()
     d.last = data
    func (d *SensorData) Get() float64 {
     defer mu.RUnlock()
     return d.last

    This uses a RWMutex which means many things can read from it at the same time while only a single thing can write. It will store a single entry much like you said.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?



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