On 12 March 2010 23:26, Scott sbuck...@gmail.com wrote:
How do I write a function 'bit' that converts an integer to binary
representation:
(bit 0) - 2r0
(bit 1) - 2r1
(bit 2) - 2r10
(bit 3) - 2r11
I understand that you want a way to obtain a string representation of
a number in binary. I
java to the rescue!
Thanks to all for your suggestions
Scott
On Mar 13, 3:45 pm, Michał Marczyk michal.marc...@gmail.com wrote:
On 12 March 2010 23:26, Scott sbuck...@gmail.com wrote:
How do I write a function 'bit' that converts an integer to binary
representation:
(bit 0) - 2r0
Two questions
How do I write a function 'bit' that converts an integer to binary
representation:
(bit 0) - 2r0
(bit 1) - 2r1
(bit 2) - 2r10
(bit 3) - 2r11
.
.
.
As well, as function 'bit-concat' with the following behavior:
(bit-concat 2r1 2r00) - 2r100
(bit-concat 2r0 2r00) - 2r000
Whenever you use the 2r0 format, the reader automatically converts it to
its base-10 Integer value. This transformation happens at the reader level
right now -- check out the 'matchNumber' method in LispReader.java for
details. So (as far as I can tell) this means that there is no standalone
uh, you are confusing representation of the thing with the thing.
Integers don't have bases, bases are used when displaying them. The
reader does not convert a 2r0 to a base-10 Integer value because
there is no such thing.
On Fri, Mar 12, 2010 at 4:23 PM, Brendan Ribera
brendan.rib...@gmail.com
Yes, yes - that's what I mean. Things get a little muddled on Friday
afternoon. The reader converts the representation, and there's not a
fast/easy way to get the original representation back and manipulate it.
On Mar 12, 2010, at 4:53 PM, Kevin Downey redc...@gmail.com wrote:
uh, you are
