Bitshift with different endianess

Hello!

Consider this code:

Code:

unsigned int i = 1;
i <<= 8;
printf("%d\n", i);

If the processor operates with little endianess, the code will print 256. But what if it has big endianess, will it print 0?

1 << 8 will be 256 in all systems.

The byte-order (endianness) is only relevant when the processor stores the data in memory. The internal structure of a CPU register has all bits in a logical order from 0..n, in a logical right to left order. This register may be stored as 4 bytes in either big or little endian order - but shifting it 8 bits will still shuffle the entire number 8 bits to the left (or right), because this happens internally in the CPU, not on the memory outside the CPU.

--
Mats

Ok, but this example:

Code:

char a[4];
a[0] = 0;
a[1] = 1;
a[2] = 2;
a[3] = 3;
printf("%.8X", *(int*)a);

will print 00010203 if the processor uses little endianess, but 03020100 if the processor uses big endianess? Or have I got it wrong?

Yes (unless I mixed up which endianess is what).

Originally Posted by TriKri

... will print 00010203 if the processor uses little endianess, but 03020100 if the processor uses big endianess? Or have I got it wrong?

In fact, it's the inverse, if the processor is little endian, it will print 03020100, and if it's big endian, it will print 00010203.

Of course, we are also assuming that the processor is adressing bytes, and not something more "exotic", e.g. 16-bits words.

in other words, as long as you don't reinterpret the bit pattern as some other type then whatever manipulations/transformations you perform will be portable. otherwise, you are treading the waters of undefined behavior.