bitget() and MSB/LSB

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David C
David C el 28 de Sept. de 2011
Hi,
I would like to clarify the functionality of bitget(). In the documentation, it is noted that "The output of bitget is independent of the endian settings of the computer you are using."
Question: Would bitget(binaryVector,1) always return the LSB of binaryVector, regardless of whether my machine is big-endian or little-endian?
Thanks, David

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Jan
Jan el 28 de Sept. de 2011
Yes. BITGET is not influenced by the endianess of the machine. Therefore bitget(x, 1) and mod(x, 2) are equivalent, if x is a non-negative integer.
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David C
David C el 2 de Oct. de 2011
Thanks Jan!

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