Question

Problem: I cannot understand the number 256 (2^8) in the extract of the IBM article:

On the other hand, if it’s a
big-endian system, the high byte is 1
and the value of x is 256.

Assume each element in an array consumes 4 bites, then the processor should read somehow: 1000 0000. If it is a big endian, it is 0001 0000 because endianness does not affect bits inside bytes. [2] Contradiction to the 256 in the article!?

Question: Why is the number 256_dec (=1000 0000_bin) and not 32_dec (=0001 0000_bin)?

[2] Endian issues do not affect sequences that have single bytes, because “byte” is considered an atomic unit from a storage point of view.

Answer 1

Because a byte is 8 bits, not 4. The 9th least significant bit in an unsigned int will have value 2^(9-1)=256. (the least significant has value 2^(1-1)=1).

From the IBM article:

unsigned char endian[2] = {1, 0};
short x;

x = *(short *) endian;

They’re correct; the value is (short)256 on big-endian, or (short)1 on little-endian.

Writing out the bits, it’s an array of {00000001_{base2}, 00000000_{base2}}. Big endian would interpret that bit array read left to right; little endian would swap the two bytes.