Variable size/Set
You are encouraged to solve this task according to the task description, using any language you may know.
Demonstrate how to specify the minimum size of a variable or a data type.
Ada
type Response is (Yes, No); -- Definition of an enumeration type with two values for Response'Size use 1; -- Setting the size of Response to 1 bit, rather than the default single byte size
D
In D, any variables of static array of zero length has a size of zero. But such data is useless, as no base type element can be accessed.
typedef long[0] zeroLength ; writefln(zeroLength.sizeof) ; // print 0
NOTE: a dynamic array variable's size is always 8 bytes, 4(32-bit) for length and 4 for a reference pointer of the actual storage somewhere in runtime memory.
The proper candidates of minimum size variable are empty structure, 1-byte size data type variable (include byte, ubyte, char and bool
), and void, they all occupy 1 byte.
byte b ; ubyte ub ; char c ; bool t ;
bool
is logically 1-bit size, but it actually occupy 1 byte.
void
can't be declared alone, and has no sizeof
property, but it can be deduced that it has 1 byte size, eg.
byte[] b ; // byte has sizeof 1 bytes b.length = 8 ; // b now should be 8*1 = 8 bytes of size b[] = cast(byte[]) new void[8] ; writefln("byte.sizeof * 8 match void.sizeof * 8"); b.length = 16 ; // b now should be 16*1 = 16 bytes b[] = cast(byte[]) new void[8] ; // any length of void[] other than 16 gives error writefln("byte.sizeof * 16 match void.sizeof * 8");
when runs, gives
byte.sizeof * 8 match void.sizeof * 8 Error: lengths don't match for array copy,16 = 8
An empty structure is logically zero size, but still occupy 1 byte.
struct Empty { } writefln(Empty.sizeof) ; // print 1
Perl
I suppose you could use vec() or similar to twiddle a single bit. The thing is, as soon as you store this in a variable, the SV (the underlying C implementation of the most simple data type) already takes a couple dozen of bytes.
In Perl, memory is readily and happily traded for expressiveness and ease of use.