Thread: bitset + memcpy

Nothing is reversed. Take the difference between English and Hebrew writing systems. English goes left to right, Hebrew goes right to left. This doesn't mean that one of them is "reversed," it means they are different. To say that Hebrew, for instance, is written "backwards" is no more justifiable than saying English is written backwards.

That's jsut splitting hairs. What I meant was that, comparing an array with n elements to a bitset with nelements, the nth element is the last one in the array, but the first (leftmost?) one in the bitset.

There is ONLY "most signficant" and "least significant."

Fair enough - on a little endian system, the the MSB on the left or the right?

Originally Posted by KBriggs

That's jsut splitting hairs. What I meant was that, comparing an array with n elements to a bitset with nelements, the nth element is the last one in the array, but the first (leftmost?) one in the bitset.



Fair enough - on a little endian system, the the MSB on the left or the right?

>> That's jsut splitting hairs.

No, it was concise. To say that endianness is "reversed" is misleading and incorrect. The bytes are swapped - nothing more.

>> Fair enough - on a little endian system, the the MSB on the left or the right?

Strictly speaking, neither! In programming, "left" and "right" generally refer to bit positions, and are endian-neutral terms. Anyway, on LE systems the least significant byte comes first (in the sense that if you were to reconstruct it from a stream of bytes, the first byte processed would be the lowest byte of the word).

>> is there a way to use memcpy to copy the bits of say an integer into a certain place in a bitset?

As a rule of thumb, using memcpy/memset or otherwise messing with bits should never be used on non-simple types (eg: most classes) since it can often corrupt important state information.