Golang - how this func is called - go

Question

I'm working with this sample code but cannot understand how this function is called and which parameter is belong.
go func(r []string) {
processData(r)
ch <- r
}(record)

Answer 1

function closures:
Go functions may be closures. A closure is a function value that
references variables from outside its body. The function may access
and assign to the referenced variables; in this sense the function is
"bound" to the variables.
To Understand this:
go func(r []string) {
processData(r)
ch <- r
}(record)
let's first declare this function:
func routine(r []string) {
processData(r)
ch <- r
}
and this global variable:
var ch = make(chan []string)
Now you may call it:
go routine(record)
this calls function named routine with input parameter named record as a goroutine.
And see: https://gobyexample.com/goroutines
Try it on The Go Playground:
package main
import (
"encoding/csv"
"flag"
"fmt"
"io"
"os"
"strings"
"time"
)
func routine(r []string) {
processData(r)
ch <- r
}
var ch = make(chan []string)
func main() {
start := time.Now()
flag.Parse()
fmt.Print(strings.Join(flag.Args(), "\n"))
if *filename == "REQUIRED" {
return
}
csvfile, err := os.Open(*filename)
if err != nil {
fmt.Println(err)
return
}
defer csvfile.Close()
reader := csv.NewReader(csvfile)
i := 0
for {
record, err := reader.Read()
if err == io.EOF {
break
} else if err != nil {
fmt.Println(err)
return
}
i++
go routine(record)
fmt.Printf("go %d %s\n", i, record)
}
for ; i >= 0; i-- {
fmt.Printf("<- %d %s\n", i, <-ch)
}
fmt.Printf("\n%2fs", time.Since(start).Seconds())
}
func processData([]string) {
time.Sleep(10 * time.Millisecond)
}
var filename = flag.String("f", "REQUIRED", "source CSV file")
var numChannels = flag.Int("c", 4, "num of parallel channels")
//var bufferedChannels = flag.Bool("b", false, "enable buffered channels")