On 01/24/2010 05:16 PM, Antti Palosaari wrote:
quoted text
>> +    af9015_config.eeprom_sum = 0;
>> +    for (reg = 0; reg<  eeprom_size / sizeof(u32); reg++) {
>> +        af9015_config.eeprom_sum *= GOLDEN_RATIO_PRIME_32;
>> +        af9015_config.eeprom_sum += le32_to_cpu(((u32 *)eeprom)[reg]);
>> +    }
>> +
>> +    deb_info("%s: eeprom sum=%.8x\n", __func__,
>> af9015_config.eeprom_sum);
> 
> Does this sum contain all 256 bytes from EEPROM? 256/4 is 64.

Yes it does. It is computed as a hashed sum of 32-bit numbers (4 bytes)
-- speed (does not matter) and larger space of hashes. Hence the
division by 4. The cast does the trick: ((u32 *)eeprom)[reg] -- reg
index is on a 4-byte basis.

regards,
-- 
js
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