Low Byte & High Byte

Hello all

I had a question regarding a Low byte and high Byte to an INT.

Ex. say you have a binary 8 bit low byte representation of 10110111 which equates to 183 decimal value and then a high byte representation 10001111 which equates to 143 decimal, what is the theoretical way of "combining" the low byte and high byte in "binary" then reading the value as one (word). I know in logix 5000 their are ways of combining these two, but that is not the method I'm looking for, I'm specifically looking for the method we learned in school how to equate this. I just can't remember how to do it. Thanks in advance

bit shift the high to the left 8 bits and then do an OR ??

@ peter that works, but do you have a more universal approach to doing this, because some processor don't have shift registers, i remember their was a way to multiply by 256 or something, i just don't remember.

You shift bits left by multiplying by 2ⁿ where n is the number of bits you want to shift. An 8 bit shift is 256.

With a CPT statement it is

CPT DestinationTag (HigByteTag*256)+LowByteTag

@ TConnolly thanks, that was the approach i was looking for, followed with that you included a "+" operator in your CPT instructin. So my question follows is that an addition of the resultant binary numbers or an "OR" of the binary numbers to get the low and high byte as a word? Unless we are taking the low and high bytes as decimals?? please explain.

In this case it doesn't matter if you OR or ADD. After the shift the lower 8 bits are all 0, so there are no bit carries.

@ Tconnolly thanks