Getting 32 bit words out of 64-bit values in C/C++ and not worrying about endianness

Question

It's my understanding that in C/C++ bitwise operators are supposed to be endian independent and behave the way you expect. I want to make sure that I'm truly getting the most significant and least significant words out of a 64-bit value and not worry about endianness of the machine. Here's an example:

uint64_t temp;
uint32_t msw, lsw;
msw = (temp & 0xFFFFFFFF00000000) >> 32;
lsw = temp & 0x00000000FFFFFFFF;

Will this work?

Answer 1

6.5.7 Bitwise shift operators

4 The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2E2, reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

So, yes -- guranteed by the standard.

Answer 2

It will work, but the strange propensity of some authors for doing bit-masking before bit-shifting always puzzled me.

In my opinion, a much more elegant approach would be the one that does the shift first

msw = (temp >> 32) & 0xFFFFFFFF; 
lsw = temp & 0xFFFFFFFF; 

at least because it uses the same "magic" bit-mask constant every time.

Now, if your target type is unsigned and already has the desired bit-width, masking becomes completely unnecesary

msw = temp >> 32; 
lsw = temp;