按位运算符和“ endianness”

You haven't specified a language but usually, programming languages such as C abstract endianness away in bitwise operations. So no, it doesn't matter in bitwise operations.

Endianness only matters for layout of data in memory. As soon as data is loaded by the processor to be operated on, endianness is completely irrelevent. Shifts, bitwise operations, and so on perform as you would expect (data logically laid out as low-order bit to high) regardless of endianness.

The bitwise operators abstract away the endianness. For example, the >> operator always shifts the bits towards the least significant digit. However, this doesn't mean you are safe to completely ignore endianness when using them, for example when dealing with individual bytes in a larger structure you cannot always assume that they will fall in the same place.

short temp = 0x1234;
temp = temp >> 8;


// on little endian, c will be 0x12, on big endian, it will be 0x0
char c=((char*)&temp)[0];

To clarify, I am not in basic disagreement with the other answers here. The point I am trying to make is to emphasise that although the bitwise operators are essentially endian neutral, you cannot ignore the effect of endianess in your code, especially when combined with other operators.

As others have mentioned, shifts are defined by the C language specification and are independent of endianness, but the implementation of a right shift may vary depending on iff the architecture uses one's complement or two's complement arithmetic.

It depends. Without casting the number into a new type, you can treat the endianness transparently.

However, if your operation involves some new type casting, then use your caution.

For example, if you want right shift some bits and cast (explicitly or not) to a new type, endianness matters!

To test your endianness, you can simply cast an int into a char:

int i = 1;


char *ptr;


...


ptr = (char *) &i;  //Cast it here


return  (*ptr);