maths programming problem

hope this is the right place to put this.

i'm writing some software, and i have encountered a little maths problem that slowing me down atm.


i have a field that is like a value in chmod or user level in HL adminmod, where a single number stores multiple states. the states are stored in the number as 2 to the power of the number option required.

eg.

2 = state 1
4 = state 2
8 = state 3
16 = state 4
32 = state 5
...
2^x = state x


to store the state 2 and 4 the single number whould equal 4 + 16 = 20

later in the system, i need to recover what states where selected. This can be done by trying to subtract the greatest possible option first, and continue down until all values for x in 2^x are covered. And the set of x's that where 2^x could be subtracted successfully from the stored number, are the states that where selected.

But i was wondering if there is a shortcut to this, to find if a particular state x (2^x value) is or isnt stored the single number.

i have hunches that y ^ 1/x (where y=single number, x=state) but i become lost when the single number contains a sum of multiple 2^x. logs or something else must surely help somehow :(

I have tried searching for an answer on the web, and finding it hard to isolate.

any help would be great.


i easily muddle myself in maths if i dont stay in regular practice (which i rarely do).

The easiest way to retrieve the values is just to mask them with an AND:

state 1 = value AND 2
state 2 = value AND 4
state 3 = value AND 8
etc

You didn't mention what language you're using, so you'll probably have to translate as appropriate.

Cheers,
Martin.

Not quite....the Graphics and Programming forum is the appropriate place, so I'm moving it to there...

Cheers,
Martin.

An elegant way to do this would be to use a bitmask for the state value.

Code:

int state1 = 0, state2 = 0, states=0;
state1 |= (1 << 2);
state2 |= (1 << 4);
states = state1 | state2;
printf("Mask = %d\n", states);

So, that means that state1 is equal to 1 bitshifted left 2 places (or in binary, 0010) which is 4, state2 is bitshifted left 4 places (1000) which is 16, and state3 is the logical OR of those two (1010) which is 20. Now all you have to do to check what values you have in there is to do what mpot stated above.

Code:

if(states & state1) 
 //do some other code

The word state was used way too much in that post.

thanks, that is prefect!

((im writing it in perl, saving to mysql (through dbi). and php will be the interface to the user. ))


yeah, sorry i wrote it at about 4am i think.

was trying to be clear between refering a state (or option) that was to be stored in the target number, and not to mixed them up in my explaination.


thanks again.