Question

Can someone help me understand whats going on with this code. It looks like it is making an integer from an array of bits. Im not sure how its doing that. Why is there a bitwise & operation on OxFF? Inst this just going to produce the same result?

//first take the first  4 bytes read out of the socket into an array and
//make them a 32 bit integer

        long ltemp =0;
        long ltemp2 = 0;
        ltemp  = ltemp | (unsigned char)(analog_val_ptr[0] & 0xff);
        ltemp  = ltemp << 24;
        ltemp2 = ltemp2 | (unsigned char)(analog_val_ptr[1] & 0xff);
        ltemp2 = ltemp2 << 16;
        ltemp = ltemp2 | ltemp;
        ltemp2 =0;
        ltemp2 = ltemp2 | (unsigned char)(analog_val_ptr[2] & 0xff);
        ltemp2 = ltemp2 << 8;
        ltemp = ltemp2 | ltemp;
        ltemp  =  ltemp | (unsigned char)(analog_val_ptr[3] & 0xff);

///then convert that integer into a float, passing

Answer 1

That’s a very long-winded way of just converting four 8-bit bytes into a 32-bit long.

The anding with 0xff is just ensuring that only the lower 8 bits of each value are used (0xff == binary 11111111).

The bit-shifting (in multiples of 8) is just to get each character into the right position.

The whole thing could be replaced with something like:

unsigned long ltemp  = (unsigned char)(analog_val_ptr[0] & 0xff);
ltemp = (ltemp << 8) | (unsigned char)(analog_val_ptr[1] & 0xff);
ltemp = (ltemp << 8) | (unsigned char)(analog_val_ptr[2] & 0xff);
ltemp = (ltemp << 8) | (unsigned char)(analog_val_ptr[3] & 0xff);

Or, alternatively (and assuming they’re available), use the correct tools for the job, specifically htonl() and ntohl().