igb: fix bit_shift to be in [1..8] range

In igb_hash_mc_addr() the expression:
        "mc_addr[4] >> 8 - bit_shift", right shifting "mc_addr[4]"
shift by more than 7 bits always yields zero, so hash becomes not so different.
Add initialization with bit_shift = 1 and add a loop condition to ensure
bit_shift will be always in [1..8] range.

Fixes: 9d5c824 ("igb: PCI-Express 82575 Gigabit Ethernet driver")
Signed-off-by: Aleksandr Loktionov <aleksandr.loktionov@intel.com>
Tested-by: Pucha Himasekhar Reddy <himasekharx.reddy.pucha@intel.com> (A Contingent worker at Intel)
Signed-off-by: Tony Nguyen <anthony.l.nguyen@intel.com>
Signed-off-by: David S. Miller <davem@davemloft.net>

Author: aloktion

drivers/net/ethernet/intel/igb/e1000_mac.c

@@ -426,15 +426,15 @@ void igb_mta_set(struct e1000_hw *hw, u32 hash_value)
 static u32 igb_hash_mc_addr(struct e1000_hw *hw, u8 *mc_addr)
 {
 	u32 hash_value, hash_mask;
-	u8 bit_shift = 0;
+	u8 bit_shift = 1;
 
 	/* Register count multiplied by bits per register */
 	hash_mask = (hw->mac.mta_reg_count * 32) - 1;
 
 	/* For a mc_filter_type of 0, bit_shift is the number of left-shifts
 	 * where 0xFF would still fall within the hash mask.
 	 */
-	while (hash_mask >> bit_shift != 0xFF)
+	while (hash_mask >> bit_shift != 0xFF && bit_shift < 4)
 		bit_shift++;
 
 	/* The portion of the address that is used for the hash table