Why a GO routine like the following outputs sequences of bytes in a random order when using a buffered channel?
Here is the code to replicate the buggy behaviour, where data.csv
is a simple CSV of 1000 rows of random data (100 bytes per row approximately) plus the header row (1001 rows in total).
package main
import (
"bufio"
"os"
"time"
)
func main() {
var channelLength = 10000
var channel = make(chan []byte, channelLength)
go func() {
for c := range channel {
println(string(c))
}
}()
file, _ := os.Open("./data.csv")
scanner := bufio.NewScanner(file)
for scanner.Scan() {
channel <- scanner.Bytes()
}
<-time.After(time.Second * time.Duration(3600))
}
Here are the first 6 lines of the output as an example of what I mean for "broken output":
979,C
tharine,Vero,cveror6@blinklist.com,Female,133.153.12.53
980,Mauriz
a,Ilett,milettr7@theguardian.com,Female,226.123.252.118
981
Sher,De Laci,sdelacir8@nps.gov,Female,137.207.30.217
[...]
On the other hand, the code runs smoothly if channelLength = 0, so with an unbuffered channel (first 6 lines, again):
id,first_name,last_name,email,gender,ip_address
1,Hebert,Edgecumbe,hedgecumbe0@apple.com,Male,108.84.217.38
2,Minor,Lakes,mlakes1@marriott.com,Male,231.185.189.39
3,Faye,Spurdens,fspurdens2@oakley.com,Female,80.173.161.81
4,Kris,Proppers,kproppers3@gmpg.org,Male,10.80.182.51
5,Bronnie,Branchet,bbranchet4@squarespace.com,Male,118.117.0.5
[...]
Data is random generated.