Re: [Qemu-devel] [PATCH 4/9] hbitmap: store / restore

[Paolo Bonzini]

On 08/01/2015 22:21, John Snow wrote:
> Why are the conversions to little endian, though? Shouldn't we be
> serializing to a Big Endian format?

Because reading two 32-bit little-endian longs or a 64-bit little-endian
long gives the same value.  This is not true for big-endian.

Take the following two 32-bit values:

   0x04030201    0x08070605

representing offsets 0 and 32 in the bitmap.  Parse them back as one
64-bit little endian value:

   0x0807060504030201 = 0x04030201 + 0x08070605 * 2^32

Now parse them as as one 64-bit big endian value:

   0x0403020108070605 => 0x08070605 + 0x04030201 * 2^32

so the values are swapped.

>> +void hbitmap_restore_data(HBitmap *hb, uint8_t *buf,
>> +                          uint64_t start, uint64_t count)
>> +{
>> +    uint64_t last = start + count - 1;
>> +    unsigned long *in = (unsigned long *)buf;
>> +
>> +    if (count == 0) {
>> +        return;
>> +    }
>> +
>> +    start = (start >> hb->granularity) >> BITS_PER_LEVEL;
>> +    last = (last >> hb->granularity) >> BITS_PER_LEVEL;
>> +    count = last - start + 1;
>> +
>> +#ifdef __BIG_ENDIAN_BITFIELD
>> +    for (i = start; i <= last; ++i) {
>> +        hb->levels[HBITMAP_LEVELS - 1][i] =
>> +            (BITS_PER_LONG == 32 ? be32_to_cpu(in[i]) :
>> be64_to_cpu(in[i]));
>> +    }
>> +#else
>> +    memcpy(&hb->levels[HBITMAP_LEVELS - 1][start], in,
>> +           count * sizeof(unsigned long));
>> +#endif
>> +}
>> +
> 
> ...? We're storing as LE but restoring from BE? I'm confused.
> 
> I'm also not clear on the __BIG_ENDIAN_BITFIELD macro. Why do we want to
> pack differently based on how C-bitfields are packed by the compiler? I
> don't think that has any impact on how longs are stored (always in the
> host native format.)

I agree.

Paolo