How to create a bit array in common lisp?

Question

After googling for about an hour, I have to confess, that while I find a lot of documentation about functions operating on bit arrays, I cannot find a single reference on how to actually create a bit array.

Right now, it seems to me that either, some arrays with other element types can be handled as bit arrays OR that one could use (make-array:element-type (???)) where I could not find any explanation as to what to put where I wrote the "???".

So, while it is probably obvious to anyone else, I have no idea how to create a bit array. I know about how to write a literal bit array - but if I need a bit array with, say 2^16 bits - how would I do it?

Answer 1

You are right about using make-array , just use 'bit as the element type. Try (make-array initial-size:element-type 'bit) . The symbol BIT names the bit type and could be replaced with any other type specifier to make an array holding objects of that type. In this example initial-size is just a variable holding a whole number.

Answer 2

Another way to create a bit vector:

> (make-sequence '(vector bit) 10)
#*0000000000

Answer 3

There's also a literal syntax using the #* reader macro, and notice that the concrete type may differ between using make-array and make-sequence , though I am not sure if performance may be different depending on that...

Tested with SBCL:

CL-USER> (defvar arr (make-array 10 :element-type 'bit :fill-pointer 0))
ARR

CL-USER> (type-of arr)
(VECTOR T 10)

CL-USER> (defvar arr3 (make-sequence '(vector bit) 10))
ARR3

CL-USER> (type-of arr3)
(SIMPLE-BIT-VECTOR 10)

CL-USER> (type-of #*0101010100)
(SIMPLE-BIT-VECTOR 10)

Answer 4

How about this:

• (setq x 10)
• 10
• (setq y (read-from-string (format nil "#*~7,'0b" x)))
• #*0001010
• The 7 is an arbitrary length, which could be set by
• (setq z 8)
• 8
• (setq y (read-from-string (format nil (concatenate 'string "#*~" (write-to-string z) ",'0b") x)))
• #*00001010
• A large bit array may be better handled as an unsigned integer.