>>> @@ -398,6 +397,7 @@ static int xhci_plat_remove(struct platform_device *dev) >>> clk_disable_unprepare(clk); >>> clk_disable_unprepare(reg_clk); >>> + usb_phy_shutdown(hcd->usb_phy); >>> usb_put_hcd(hcd); Is it ok to put usb_phy_shutdown before usb_put_hcd(hcd)? hcd is released at usb_put_hcd. UNISOC DWC3 phy is not divided USB 2.0/3.0 phy clearly. Yes, it's UNISOC's issue. It UNISOC's dtsi: phys = <&ssphy>, <&ssphy>; If to shutdown phy too earlier, it will cost 10s timeout to do xhci_reset. usb_remmove_hcd --> usb_stop_hcd --> xhci_stop --> xhci_reset --> xhci_handshake(&xhci->op_regs->command, CMD_RESET, 0, 10 * 1000 *1000) I want to know this change is acceptable or not? hcd->usb_phy = devm_usb_get_phy_by_phandle(sysdev, "usb-phy", 0); Why in xhci_plat_remove, just to shutdown "usb-phy"[0], not to shutdown "usb-phy"[1] ? Mathias Nyman <mathias.nyman@xxxxxxxxxxxxxxx> 于2022年4月22日周五 15:51写道： > > On 22.4.2022 5.10, surong pang wrote: > > shared_hcd might be freed already here, but hcd should not be freed > > here, because "usb_put_hcd(hcd)" is called later. > > To me it still looks like this patch calls usb_phy_shutdown(hcd->usb_phy) _after_ > usb_put_hcd(hcd): > > >>> @@ -398,6 +397,7 @@ static int xhci_plat_remove(struct platform_device *dev) > >>> clk_disable_unprepare(clk); > >>> clk_disable_unprepare(reg_clk); > >>> usb_put_hcd(hcd); > >>> + usb_phy_shutdown(hcd->usb_phy); > > > shared hcd was freed even earlier, before disabling the clocks. > > Thanks > Mathias
