Re: [PATCH] staging: vt6656: Use BIT_ULL() macro instead of bit shift operation

Malcolm Priestley
2020-03-08T19:22:14Z

>>> */
>>> #undef __NO_VERSION__
>>>
>>> +#include <linux/bits.h>
>>> #include <linux/etherdevice.h>
>>> #include <linux/file.h>
>>> #include "device.h"
>>> @@ -802,8 +803,7 @@ static u64 vnt_prepare_multicast(struct ieee80211_hw *hw,
>>>
>>> netdev_hw_addr_list_for_each(ha, mc_list) {
>>> bit_nr = ether_crc(ETH_ALEN, ha->addr) >> 26;
>>> -
>>> - mc_filter |= 1ULL << (bit_nr & 0x3f);
>>> + mc_filter |= BIT_ULL(bit_nr);
>>
>> Are you sure this does the same thing? You are not masking off bit_nr
>> anymore, why not?
>
> My reasons are exposed below:
>
> The ether_crc function returns an u32 type (unsigned of 32 bits). Then the right
> shift operand discards the 26 lsb bits (the bits shifted off the right side are
> discarded). The 6 msb bits of the u32 returned by the ether_crc function are
> positioned in bit 5 to bit 0 of the variable bit_nr. Due to the right shift
> happens over an unsigned type, the 26 new bits added on the left side will be 0.
>
> In summary, after the right bit shift operation we obtain in the variable bit_nr
> (unsigned of 32 bits) the value represented by the 6 msb bits of the value
> returned by the ether_crc function. So, only the 6 lsb bits of the variable
> bit_nr are important. The 26 msb bits of this variable are 0.
>
> In this situation, the "and" operation with the mask 0x3f (mask of 6 lsb bits)
> is unnecessary due to its purpose is to reset (set to 0 value) the 26 msb bits
> that are yet 0.

The mask is only there out of legacy originally it was 31(0x1f) and the
bit_nr spread across two mc_filter u32 arrays.

The mask is not needed now it is u64.

The patch is fine.

Regards

Malcolm