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Endianness Concern with Increment

Posted on 2007-11-25
Medium Priority
Last Modified: 2010-04-15
I often use 64-bit integers to store aggregate values to be stored on disk.  

Suppose I use the last 8-bits of a 64-bit integer to store some value from 0 to 255.  Then suppose I want to increment that value, without affecting the rest of the 64 bit integer.

This could be done in 2 ways.  I could use bitwise operations like this:

val = (val & (0xFFUL << 56)) + (1UL << 56) | (val &  0xFFFFFFFFFFFFFFUL);

Or I could simply address the last byte directly and increment it:

++(*(  ((unsigned char*) &val) + 7));

The latter method is probably more efficient, but is it portable and safe?  As far as I can see, this should be fine on both big and little endian computers, and also should not cause any alignment issues because I am addressing a single byte.  Still, I'm not sure.  Am I correct?  Is the latter method portable/safe?
Question by:chsalvia
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LVL 40

Expert Comment

ID: 20347395
Remember, you have byte Endianness as well as bit Endianness!
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Expert Comment

ID: 20347397
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20347404
>>  Or I could simply address the last byte directly and increment it:
>>  ++(*(  ((unsigned char*) &val) + 7));
Casting an integer as an array pointer is not portable considering the big/little endian organization, it will return the LSB or the MSB depending on platform.
Better you use the first method.

LVL 55

Accepted Solution

Jaime Olivares earned 1000 total points
ID: 20347414
indeed you can increment by simply do this:
val += 0x0100000000000000UL;

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