usb: usbip: vudc: fix left shift overflow

In v_recv_cmd_submit(), urb_p->urb->pipe has the type unsigned int
(which is 32-bit long on x86_64) but 11<<30 results in a 34-bit integer.
Therefore the 2 leading bits are truncated and

    urb_p->urb->pipe &= ~(11 << 30);

has the same meaning as

    urb_p->urb->pipe &= ~(3 << 30);

This second statement seems to be how the code was intended to be
written, as PIPE_ constants have values between 0 and 3.

The overflow has been detected with a clang warning:

    drivers/usb/usbip/vudc_rx.c:145:27: warning: signed shift result
    (0x2C0000000) requires 35 bits to represent, but 'int' only has 32
    bits [-Wshift-overflow]
            urb_p->urb->pipe &= ~(11 << 30);
                                  ~~ ^  ~~

Fixes: 79c02cb1fd5c ("usbip: vudc: Add vudc_rx")
Signed-off-by: Nicolas Iooss <nicolas.iooss_linux@m4x.org>
Signed-off-by: Greg Kroah-Hartman <gregkh@linuxfoundation.org>

drivers/usb/usbip/vudc_rx.c

@@ -142,7 +142,7 @@ static int v_recv_cmd_submit(struct vudc *udc,
 	urb_p->urb->status = -EINPROGRESS;
 
 	/* FIXME: more pipe setup to please usbip_common */
-	urb_p->urb->pipe &= ~(11 << 30);
+	urb_p->urb->pipe &= ~(3 << 30);
 	switch (urb_p->ep->type) {
 	case USB_ENDPOINT_XFER_BULK:
 		urb_p->urb->pipe |= (PIPE_BULK << 30);