add support for arrow/up/down/left/right

operation.go

@@ -75,6 +75,7 @@ func (l *Operation) ioloop() {
 		case MetaBackspace:
 			l.buf.BackEscapeWord()
 		case CharEnter, CharEnter2:
+			l.buf.MoveToLineEnd()
 			l.buf.WriteRune('\n')
 			data := l.buf.Reset()
 			data = data[:len(data)-1] // trim \n

terminal.go

@@ -18,11 +18,12 @@ const (
 )
 
 const (
-	KeyPrevChar  = 0x2
-	KeyInterrupt = 0x3
-	KeyNextChar  = 0x6
-	KeyDelete    = 0x4
-	KeyEsc       = 0x1b
+	KeyPrevChar  = 2
+	KeyInterrupt = 3
+	KeyNextChar  = 6
+	KeyDelete    = 4
+	KeyEsc       = 27
+	KeyEscapeEx  = 91
 )
 
 type Terminal struct {
@@ -67,16 +68,26 @@ func (t *Terminal) ReadRune() rune {
 
 func (t *Terminal) ioloop() {
 	buf := bufio.NewReader(os.Stdin)
-	prefix := false
+	isEscape := false
+	isEscapeEx := false
 	for {
 		r, _, err := buf.ReadRune()
+		Debug(r, isEscape, isEscapeEx)
 		if err != nil {
 			break
 		}
 
-		if prefix {
-			prefix = false
-			r = prefixKey(r)
+		if isEscape {
+			isEscape = false
+			if r == KeyEscapeEx {
+				isEscapeEx = true
+				continue
+			}
+			r = escapeKey(r)
+		} else if isEscapeEx {
+			isEscapeEx = false
+			r = escapeExKey(r)
+			Debug("r:", r)
 		}
 
 		if IsPrintable(r) || r < 0 {
@@ -88,7 +99,7 @@ func (t *Terminal) ioloop() {
 			t.outchan <- r
 			goto exit
 		case KeyEsc:
-			prefix = true
+			isEscape = true
 		case CharEnter, CharEnter2, KeyPrevChar, KeyNextChar, KeyDelete:
 			fallthrough
 		case CharLineEnd, CharLineStart, CharNext, CharPrev:

utils.go

@@ -28,7 +28,21 @@ func IsPrintable(key rune) bool {
 	return key >= 32 && !isInSurrogateArea
 }
 
-func prefixKey(r rune) rune {
+func escapeExKey(r rune) rune {
+	switch r {
+	case 'D':
+		r = CharBackward
+	case 'C':
+		r = CharForward
+	case 'A':
+		r = CharPrev
+	case 'B':
+		r = CharNext
+	}
+	return r
+}
+
+func escapeKey(r rune) rune {
 	switch r {
 	case 'b':
 		r = MetaPrev
@@ -64,7 +78,6 @@ func getWidth() int {
 		uintptr(syscall.TIOCGWINSZ),
 		uintptr(unsafe.Pointer(ws)))
 
-	Debug(ws)
 	if int(retCode) == -1 {
 		panic(errno)
 	}