[rkomar]

"<<8" shifts left by 8 bits, which is the same as multiplying by 256. So, the function returns 256*c[0] + c[1], i.e. the program is interpreting those two bytes as being uint16_t in big endian order.

The registers in which the math takes place are probably at least 32 bits wide, so the addition of the temporary c[0]<<8 and c[1] values is done without truncating back to 8-bits (size of unsigned char) before it is returned as an unsigned int.