Suppose I added uint16_t x = *(uint16_t*)a << 8; Would the bit pattern for x on the little endian system be 00000000 10000000 and 00000010 00000000 on the big endian system? If that's the case would that not support the Linux article?
Suppose I added uint16_t x = *(uint16_t*)a << 8; Would the bit pattern for x on the little endian system be 00000000 10000000 and 00000010 00000000 on the big endian system? If that's the case would that not support the Linux article?