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Date: Mon, 21 Jun 2010 14:40:06 +0300
From: "Michael S. Tsirkin" <mst@redhat.com>
Message-ID: <20100621114006.GB10001@redhat.com>
References: <20100615034727.GT23473@valinux.co.jp>
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Subject: [Qemu-devel] Re: [PATCH] pci hotplug: make pci hotplug return value
	to caller.
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To: Isaku Yamahata <yamahata@valinux.co.jp>
Cc: qemu-devel@nongnu.org, Gerd Hoffmann <kraxel@redhat.com>

On Tue, Jun 15, 2010 at 12:47:27PM +0900, Isaku Yamahata wrote:
> Make pci hotplug callback return value to caller.
> There is no reason to discard return value.
> 
> Signed-off-by: Isaku Yamahata <yamahata@valinux.co.jp>
> ---
>  hw/pci.c |   11 +++++++----
>  1 files changed, 7 insertions(+), 4 deletions(-)
> 
> diff --git a/hw/pci.c b/hw/pci.c
> index 7787005..3777c1c 100644
> --- a/hw/pci.c
> +++ b/hw/pci.c
> @@ -1623,8 +1623,12 @@ static int pci_qdev_init(DeviceState *qdev, DeviceInfo *base)
>          pci_dev->romfile = qemu_strdup(info->romfile);
>      pci_add_option_rom(pci_dev);
>  
> -    if (qdev->hotplugged)
> -        bus->hotplug(bus->hotplug_qdev, pci_dev, 1);
> +    if (qdev->hotplugged) {
> +        rc = bus->hotplug(bus->hotplug_qdev, pci_dev, 1);
> +        if (rc != 0) {
> +            return rc;

I think we need to unregister the device (and remove option rom?),
otherwise there's a resource leak here.
Can it really fail? If not we can just make it void.


> +        }
> +    }
>      return 0;
>  }
>  
> @@ -1632,8 +1636,7 @@ static int pci_unplug_device(DeviceState *qdev)
>  {
>      PCIDevice *dev = DO_UPCAST(PCIDevice, qdev, qdev);
>  
> -    dev->bus->hotplug(dev->bus->hotplug_qdev, dev, 0);
> -    return 0;
> +    return dev->bus->hotplug(dev->bus->hotplug_qdev, dev, 0);
>  }
>  
>  void pci_qdev_register(PCIDeviceInfo *info)
> -- 
> 1.6.6.1
>