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

[Greg Kroah-Hartman]

On Sun, Mar 08, 2020 at 07:22:07PM +0000, Malcolm Priestley wrote:
> >>>   */
> >>>  #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.

Ok, then the changelog needs to be fixed up to explain all of this and
resent.

thanks,

greg k-h