I beleive int is the size of the system word; so in the case of 16-bit systems sizeof(int)==2, 32 bit systems sizeof(int)==4, and 64 bit systems sizeof(int)==8.

sizeof(char), I think, is always 1. For wide chars (wchar_t) the size is 2 bytes.

MBCS use two chars to represent one character, however sizeof(char) is still 1.

quote:
The only thing you are guaranteed is that:

sizeof(char) == 1
sizeof(char) <= sizeof(short) <= sizeof(int) <= sizeof(long)

This is correct.... and remember it well everyone. I started working on a new platform, porting some existing code, and got very unexpected results because sizeof(long) changed from 32-bits to 64-bits.

But because I had used a typedef''d type everywhere, it was easy to change only the typedef to fix it all.

(I haven''t tested it yet, but I''m guessing (long long) on my platform is 128-bit.)



---- --- -- -
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It says in my trusty Soustrup book:

1 == sizeof( char ) <= sizeof( short ) <= sizeof( int ) <= sizeof( long ).

Unfortunately, that 1 does NOT have any units. Sizeof simply returns sizes in multiple of char!
Therefore, it is NOT guaranteed that a char is one byte.
It is only guaranteed ( through another spec ) that char is AT LEAST 8 bits, a short at least 16, and a long at least 32.



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you could go:

union 32bit{unsinged long MyVal;struct{ unsigned char Byte1; unsigned char Byte2; unsigned char Byte3; unsigned char Byte4;}} 

and declare you''r 32bit thingy to be a 32bit then just move the bytes to their correct positions with simple assignments

note that using unions this way is way faster than any shift operations will ever be!
Be sure that you put the values in the right places, intel boxes have a weird way of storing words and dwords... but used correctly it works.