CODEWITHSUNDEEP
cbit-shift
user1203092
user1203092

Reputation: 41

C bitarray set value

I am trying to speed up a program with bitshifts. For doing that I am allocating some space for the bits to store.

Now I have to set the bit in the middle to 1. So for example I want:

00000010000000

to be in my memory. (The length is defined by the user on run-time).

Here is my Code:

 int *state = malloc(b); // Where 'b' is a user defnied value
 *state |= (1 << b);

But when I print out the bits I get something like this (for b = 10):

000000000010000000000000000000000000000000100000000000000000000000000000001000000000000000000000000

So it keeps placing many 1's instead of one 1 in the middle.

What am I doing wrong here?

I belived it has to do with me using int, maybe the workaround would be not using int. But I don't know how to allocate a bitarray in another way.

This is how I print the bits (requestet in comments):

 64         for(i =0; i < b; i++){
 65                 printf("%d", ((*state & (1 << i)) >> i));
 66         }

Upvotes: 0

Views: 147

Answers (1)

Jongware
Jongware

Reputation: 22457

You are mixing up bits and bytes.

To allocate b bits, you need

int *state = malloc(sizeof(int)*((b+sizeof(int)-1)/sizeof(int))); // Where 'b' is a user defnied value

To set a bit in this array, decompose into int position and bit position:

state[b/sizeof(int)] |= (1 << (b % sizeof(int)));

If you are going to work with single bits, it might be easier to use unsigned char instead of int as the base array.

Edit: just changed the allocation from (b+sizeof(int)-1)/sizeof(int) -- this calculates the required byte size. Of course you need to allocate room for int sized members, so the allocated memory size is the number of bytes times sizeof(int).

Upvotes: 6

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