> From: Jian-Hong Pan
>
> There is a mass of jobs between spin lock and unlock in the hardware
> IRQ which will occupy much time originally. To make system work more
> efficiently, this patch moves the jobs to the soft IRQ (bottom half) to
> reduce the time in hardware IRQ.
>
> Signed-off-by: Jian-Hong Pan <jian-hong@xxxxxxxxxxxx>
> ---
> v2:
> Change the spin_lock_irqsave/unlock_irqrestore to spin_lock/unlock in
> rtw_pci_interrupt_handler. Because the interrupts are already disabled
> in the hardware interrupt handler.
>
> drivers/net/wireless/realtek/rtw88/pci.c | 33 +++++++++++++++++++-----
> 1 file changed, 27 insertions(+), 6 deletions(-)
>
> diff --git a/drivers/net/wireless/realtek/rtw88/pci.c
> b/drivers/net/wireless/realtek/rtw88/pci.c
> index 00ef229552d5..0740140d7e46 100644
> --- a/drivers/net/wireless/realtek/rtw88/pci.c
> +++ b/drivers/net/wireless/realtek/rtw88/pci.c
> @@ -866,12 +866,28 @@ static irqreturn_t rtw_pci_interrupt_handler(int irq,
> void *dev)
> {
> struct rtw_dev *rtwdev = dev;
> struct rtw_pci *rtwpci = (struct rtw_pci *)rtwdev->priv;
> - u32 irq_status[4];
>
> spin_lock(&rtwpci->irq_lock);
> if (!rtwpci->irq_enabled)
> goto out;
>
> + /* disable RTW PCI interrupt to avoid more interrupts before the end of
> + * thread function
> + */
> + rtw_pci_disable_interrupt(rtwdev, rtwpci);
So basically it's to prevent back-to-back interrupts.
Nothing wrong about it, I just wondering why we don't like
back-to-back interrupts. Does it means that those interrupts
fired between irq_handler and threadfin would increase
much more time to consume them.
> +out:
> + spin_unlock(&rtwpci->irq_lock);
> +
> + return IRQ_WAKE_THREAD;
> +}
> +
> +static irqreturn_t rtw_pci_interrupt_threadfn(int irq, void *dev)
> +{
> + struct rtw_dev *rtwdev = dev;
> + struct rtw_pci *rtwpci = (struct rtw_pci *)rtwdev->priv;
> + unsigned long flags;
> + u32 irq_status[4];
> +
> rtw_pci_irq_recognized(rtwdev, rtwpci, irq_status);
>
> if (irq_status[0] & IMR_MGNTDOK)
> @@ -891,8 +907,11 @@ static irqreturn_t rtw_pci_interrupt_handler(int irq,
> void *dev)
> if (irq_status[0] & IMR_ROK)
> rtw_pci_rx_isr(rtwdev, rtwpci, RTW_RX_QUEUE_MPDU);
>
> -out:
> - spin_unlock(&rtwpci->irq_lock);
> + /* all of the jobs for this interrupt have been done */
> + spin_lock_irqsave(&rtwpci->irq_lock, flags);
I suggest to protect the ISRs. Because next patches will require
to check if the TX DMA path is empty. This means I will also add
this rtwpci->irq_lock to the TX path, and check if the skb_queue
does not have any pending SKBs not DMAed successfully.
> + if (rtw_flag_check(rtwdev, RTW_FLAG_RUNNING))
Why check the flag here? Is there any racing or something?
Otherwise it looks to break the symmetry.
> + rtw_pci_enable_interrupt(rtwdev, rtwpci);
> + spin_unlock_irqrestore(&rtwpci->irq_lock, flags);
>
> return IRQ_HANDLED;
> }
> @@ -1152,8 +1171,10 @@ static int rtw_pci_probe(struct pci_dev *pdev,
> goto err_destroy_pci;
> }
>
> - ret = request_irq(pdev->irq, &rtw_pci_interrupt_handler,
> - IRQF_SHARED, KBUILD_MODNAME, rtwdev);
> + ret = devm_request_threaded_irq(rtwdev->dev, pdev->irq,
> + rtw_pci_interrupt_handler,
> + rtw_pci_interrupt_threadfn,
> + IRQF_SHARED, KBUILD_MODNAME, rtwdev);
> if (ret) {
> ieee80211_unregister_hw(hw);
> goto err_destroy_pci;
> @@ -1192,7 +1213,7 @@ static void rtw_pci_remove(struct pci_dev *pdev)
> rtw_pci_disable_interrupt(rtwdev, rtwpci);
> rtw_pci_destroy(rtwdev, pdev);
> rtw_pci_declaim(rtwdev, pdev);
> - free_irq(rtwpci->pdev->irq, rtwdev);
> + devm_free_irq(rtwdev->dev, rtwpci->pdev->irq, rtwdev);
> rtw_core_deinit(rtwdev);
> ieee80211_free_hw(hw);
> }
> --
> 2.20.1
Yan-Hsuan
