> On Sun, Jun 11, 2023 at 01:35:17AM +0100, Daniel Golle wrote:
> > @@ -1106,14 +1105,14 @@ struct mtk_eth {
> > spinlock_t tx_irq_lock;
> > spinlock_t rx_irq_lock;
> > struct net_device dummy_dev;
> > - struct net_device *netdev[MTK_MAX_DEVS];
> > - struct mtk_mac *mac[MTK_MAX_DEVS];
> > + struct net_device **netdev;
> > + struct mtk_mac **mac;
> > int irq[3];
> > u32 msg_enable;
> > unsigned long sysclk;
> > struct regmap *ethsys;
> > struct regmap *infra;
> > - struct phylink_pcs *sgmii_pcs[MTK_MAX_DEVS];
> > + struct phylink_pcs **sgmii_pcs;
> > struct regmap *pctl;
> > bool hwlro;
> > refcount_t dma_refcnt;
>
> Is it really worth the extra allocations?
>
> There's three pointers here per device. Let's talk about modern systems,
> so that's 8 bytes each, and if MTK_MAX_DEVS was two, that's 48 bytes in
> all. If we expanded the array to allow three, that would be 72 bytes.
>
> If we allocate separately, then we're allocating 16 or 24 bytes three
> times depending on whether we want two or three of them.
>
> On arm64, I'm seeing the minimum slab size as 128 bytes, which means
> that's the minimum memory allocation. So, allocating three arrays will
> be 384 bytes in all, irrespective of whether we want two or three
> entries.
>
> That's a waste of about 5x the memory over just expanding the arrays!
ack, I agree. I will fix it.
Regards,
Lorenzo
>
> If you want to go down the route of dynamically allocating these, it
> would make better sense to combine them into a single structure that
> itself is an array, and thus requiring only one allocation. That
> reduces the wastage to about 56 bytes for three ports or 80 bytes
> for two.
>
> Thanks.
>
> --
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>
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