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Endianness Concern with Increment

Posted on 2007-11-25
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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?
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Question by:chsalvia
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4 Comments
 
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Expert Comment

by:evilrix
ID: 20347395
Remember, you have byte Endianness as well as bit Endianness!
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by:evilrix
ID: 20347397
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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.

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