在小端机器上添加两个大端值

Question

I am currently facing a problem of which I have no idea how to avoid it..
I try to process data which can be either in big endian or little endian. This is not really a problem because it always starts with a header so I can check which endian mode I have to use but during the decoding of the values there are some operations which I dont know how to implement for big endian data.
The code runs on a nVidia Tegra (Cortex-A9 based on ARMv7 architecture) which is little endian (or runs in little endian mode) but sometimes I get big endian data.
Most operations on the data are not really a problem but I dont know how to get the addition right..

Example:    D5 1B EE 96    |     96 EE 1B D5
        +   AC 84 F4 D5    | +   D5 F4 84 AC
        = 1 81 A0 E3 6B    | = 1 6C E2 A0 81

As you can see, most bytes are already correct in the result but some are not. They differ by +1 or -1 from the expected result because the addition is always made from right to left (little endian machine) and so we take the carry (if any) to the left.
In the case of the big endian addition on this little endian machine I would have to add from left to right and take the carry (if any) to the right.

My question now is, whether there is a possibility (maybe using special instructions for the processor?) to get the right result? Maybe ther are further operations I can make on the result to get rid of these +1/-1 differences which are "cheaper" than to revert both operands and also the result?

Best Regards, Tobias

Answer 1

The most logical way to do this is to simply convert the numbers to the correct endianness, then perform the calculation, then (if needed) convert back again.

You could of course use a loop to do the byte-by-byte backwards caclulation and handle the carry - but it's more complicated, and I'm pretty certain that it won't be faster either, because there are more conditionals and processors are pretty good at "byteswapping".

You should be able to use the ntohl and htons networking functions to convert the numbers.

Something like this:

int add_big_endian(int a, int b)
{
   x = ntohl(a);
   y = ntohl(b);

   z = x + y;

   return htonl(z);
}

Answer 2

You have two options: you can write two sets of code, one for each endianness, and try to keep track of what's going on where, or you can use a single internal representation and convert incoming and outgoing values appropriately. The latter is much simpler.