[PATCH v2 2/6] PCI/MSI: Factor out pci_get_msi_cap() interface
Alexander Gordeev
agordeev at redhat.com
Thu Sep 26 20:45:48 EST 2013
On Thu, Sep 26, 2013 at 09:58:53AM +0100, David Laight wrote:
> Would it be possible to do some kind of 2-stage allocation.
> In the first pass the driver would pass a minimum and desired
> number of MSI-X interrupts, but not actually be given any.
Repeated calls to msi_enable_msi/msix() is what we are trying to avoid.
> Interrupts could then be allocated after it is known how many
> are required and how many are available.
Yep, that what we are heading to. So basic pattern I see would be like this:
int foo_driver_enable_msix(struct pci_dev *pdev, int nvec)
{
...
rc = pci_msix_table_size(pdev);
if (rc < 0)
return rc;
nvec = min(nvec, rc);
if (nvec < FOO_DRIVER_MINIMUM_NVEC)
goto single_msi;
for (i = 0; i < nvec; i++)
entries[i].entry = i;
rc = pci_enable_msix(pdev, entries, nvec);
if (rc)
goto single_msi;
return 0;
single_msi:
...
}
But this will break pSeries and we might end up with something like this:
int foo_driver_enable_msix(struct pci_dev *pdev, int nvec)
{
...
rc = pci_msix_table_size(pdev);
if (rc < 0)
return rc;
nvec = min(nvec, rc);
if (nvec < FOO_DRIVER_MINIMUM_NVEC)
goto single_msi;
rc = pci_get_msix_limit(pdev, nvec);
if (rc < 0)
return rc;
nvec = min(nvec, rc);
if (nvec < FOO_DRIVER_MINIMUM_NVEC)
goto single_msi;
for (i = 0; i < nvec; i++)
entries[i].entry = i;
rc = pci_enable_msix(pdev, entries, nvec);
if (rc)
goto single_msi;
return 0;
single_msi:
...
}
> David
--
Regards,
Alexander Gordeev
agordeev at redhat.com
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