I'd like to read and process 1024 bytes at a time in my file given by filename. I don't understand how to construct the outer loop correctly, especially to accommodate the final stride in which the buffer will contain fewer than 1024 bytes
What I have tried:
fs, _ := os.Open(filename)
defer fs.Close()
n := 1024 // 1kb
buff := make([]byte, n)
for {
buff = make([]byte, n) // is this initialized correctly?
n1, err := fs.Read(buff)
if err != nil {
if err == io.EOF {
break
}
fmt.Println(err)
break
}
fmt.Println("read n1 bytes...", n1)
fmt.Println("%v", buff)
}
It has produced the following errors:
I have seen the following resources:
I'd like to read and process 1024 bytes at a time in my file given by filename.
I would expect to see something like this:
package main
import (
"bufio"
"fmt"
"io"
"os"
)
func main() {
filename := `test.file`
f, err := os.Open(filename)
if err != nil {
fmt.Println(err)
return
}
defer f.Close()
r := bufio.NewReader(f)
buf := make([]byte, 0, 1024)
for {
n, err := io.ReadFull(r, buf[:cap(buf)])
buf = buf[:n]
if err != nil {
if err == io.EOF {
break
}
if err != io.ErrUnexpectedEOF {
fmt.Println(err)
break
}
}
fmt.Println("read n bytes...", n)
// process buf
}
}
Output:
read n bytes... 1024
read n bytes... 1024
read n bytes... 1024
read n bytes... 80