When moving a soundfile from one computer to another, such as from a ``PC'' to a ``Mac'' (Intel processor to Motorola processor), the bytes in each sound sample have to be swapped. This is because Motorola processors are big endian (bytes are numbered from most-significant to least-significant in a multi-byte word) while Intel processors are little endian (bytes are numbered from least-significant to most-significant).G.4 Any Mac program that supports a soundfile format native to PCs (such as .wav files) will swap the bytes for you. You only have to worry about swapping the bytes yourself when reading raw binary soundfiles from a foreign computer, or when digging the sound samples out of an ``unsupported'' soundfile format yourself.
Since soundfiles typically contain 16 bit samples (not for any good reason, as we now know), there are only two bytes in each audio sample. Let L denote the least-significant byte, and M the most-significant byte. Then a 16-bit word is most naturally written , i.e., the most-significant byte is most naturally written to the left of the least-significant byte, analogous to the way we write binary or decimal integers. This ``most natural'' ordering is used as the byte-address ordering in big-endian processors:
Since a byte (eight bits) is the smallest addressable unit in modern day processor families, we don't have to additionally worry about reversing the bits in each byte. Bits are not given explicit ``addresses'' in memory. They are extracted by means other than simple addressing (such as masking and shifting operations, table look-up, or using specialized processor instructions).
When compiling C or C++ programs under UNIX, there may be a macro constant called BYTE_ORDER in the header file endian.h or bytesex.h. In other cases, there may be macros such as __INTEL__, __LITTLE_ENDIAN__, __BIG_ENDIAN__, or the like, which can be detected at compile time using #ifdef.
How Many Bits are Enough for Digital Audio?