question about binary numbers

[jinduy]

suppose i have an integer i = 11000001 (binary)

what part of i would be considered the higher order bits ih and what part of it would be considered the lower order bits il ? Do you just split them in half?

thanks...unfortunately i have no friends in class to turn to and this book im reading doesn't define stuff for me 🙁

 

I would like to help you, but the whole integer and notation mixture has me confused. Hopefully this bump will help you get a response from someone who understands your question. If you care to, you can elaborate though.

 

[Storm]

Originally posted by: jinduy
suppose i have an integer i = 11000001 (binary)

what part of i would be considered the higher order bits ih and what part of it would be considered the lower order bits il ? Do you just split them in half?

thanks...unfortunately i have no friends in class to turn to and this book im reading doesn't define stuff for me 🙁

What class is this for?

 

[melly]

i think the numbers are calculated from right to left, but the final number is read from left to right.

is that what you meant?

 

[jinduy]

oh thanks guys i figured it out duh me u just split them in half so i = 1001, ih = 10, il = 01 this is for my algorithms class...needa freshen up on the terminology =[

 

[wake]

Always thought the first 4 bits are the high and the last 4 bits are the low
if you had only 5 bits 000xxxxx then your low = xxxx and your high=000x
Correct me if I'm wrong

 

[RU482]

I'm thinking of starting to use a base-36 number system (zero-z) to mess with my coworkers

 

[DaveSimmons]

Originally posted by: wake
Always thought the first 4 bits are the high and the last 4 bits are the low
if you had only 5 bits 000xxxxx then your low = xxxx and your high=000x
Correct me if I'm wrong

yup, assuming an 8-bit byte (old weird mainframes and minis from 30-40 years ago had other lengths)

 

[Calin]

A bit is higher order than the bits that is right of it, and is lower order than the bits that are in his right.
If you have a 16-bit value, then you have high-order bytes and low order bytes. But what you have there is a byte (8 bits), so if you want to make a split, you'll have last four bits as low order, and first four bits as high-order

Calin

 

[notfred]

high order bits are on the left, low ordre bits are on the right.

 

[bUnMaNGo]

don't forget big endian and little endian actually it doesn't apply here but just to throw that at you to confuse you some more 😛

 

[DaveSimmons]

Originally posted by: notfred
high order bits are on the left, low ordre bits are on the right.

An easy way to remember this is that it works just like with decimal numbers.

If you see 768 decimal you know the 7 is the "biggest" number and 8 is the "littlest" number, right?

Same for binary except each digit left is multiplied by a power of 2 instead of power of 10.

 

[Calin]

Originally posted by: bUnMaNGo
don't forget big endian and little endian actually it doesn't apply here but just to throw that at you to confuse you some more 😛

Shame on me, how could I forget of big and little endian 😱

Calin