C variable smaller then 8-bit - c

Question

I'm writing C implementation of Conway's Game of Life and pretty much done with the code, but I'm wondering what is the most efficient way to storage the net in the program.
The net is two dimensional and stores whether cell (x, y) is alive (1) or dead (0). Currently I'm doing it with unsigned char like that:
struct:
typedef struct {
int rows;
int cols;
unsigned char *vec;
} net_t;
allocation:
n->vec = calloc( n->rows * n->cols, sizeof(unsigned char) );
filling:
i = ( n->cols * (x - 1) ) + (y - 1);
n->vec[i] = 1;
searching:
if( n->vec[i] == 1 )
but I don't really need 0-255 values - I only need 0 - 1, so I'm feeling that doing it like that is a waste of space, but as far as I know 8-bit char is the smallest type in C.
Is there any way to do it better?
Thanks!

Answer 1

The smallest declarable / addressable unit of memory you can address/use is a single byte, implemented as unsigned char in your case.
If you want to really save on space, you could make use of masking off individual bits in a character, or using bit fields via a union. The trade-off will be that your code will execute a bit slower, and will certainly be more complicated.
#include <stdio.h>
union both {
struct {
unsigned char b0: 1;
unsigned char b1: 1;
unsigned char b2: 1;
unsigned char b3: 1;
unsigned char b4: 1;
unsigned char b5: 1;
unsigned char b6: 1;
unsigned char b7: 1;
} bits;
unsigned char byte;
};
int main ( ) {
union both var;
var.byte = 0xAA;
if ( var.bits.b0 ) {
printf("Yes\n");
} else {
printf("No\n");
}
return 0;
}
References
Union and Bit Fields, Accessed 2014-04-07, <http://www.rightcorner.com/code/CPP/Basic/union/sample.php>
Access Bits in a Char in C, Accessed 2014-04-07, <https://stackoverflow.com/questions/8584577/access-bits-in-a-char-in-c>
Struct - Bit Field, Accessed 2014-04-07, <http://cboard.cprogramming.com/c-programming/10029-struct-bit-fields.html>

Answer 2

Unless you're working on an embedded platform, I wouldn't be too concerned about the size your net takes up by using an unsigned char to store only a 1 or 0.
To address your specific question: char is the smallest of the C data types. char, signed char, and unsigned char are all only going to take up 1 byte each.
If you want to make your code smaller you can use bitfields to decrees the amount of space you take up, but that will increase the complexity of your code.
For a simple exercise like this, I'd be more concerned about readability than size. One way you can make it more obvious what you're doing is switch to a bool instead of a char.
#include <stdbool.h>
typedef struct {
int rows;
int cols;
bool *vec;
} net_t;
You can then use true and false which, IMO, will make your code much easier to read and understand when all you need is 1 and 0.
It will take up at least as much space as the way you're doing it now, but like I said, consider what's really important in the program you're writing for the platform you're writing it for... it's probably not the size.

Answer 3

The smallest type on C as i know are the char (-128, 127), signed char (-128, 127), unsigned char (0, 255) types, all of them takes a whole byte, so if you are storing multiple bits values on different variables, you can instead use an unsigned char as a group of bits.
unsigned char lives = 128;
At this moment, lives have a 128 decimal value, which it's 10000000 in binary, so now you can use a bitwise operator to get a single value from this variable (like an array of bits)
if((lives >> 7) == 1) {
//This code will run if the 8 bit from right to left (decimal 128) it's true
}
It's a little complex, but finally you'll end up with a bit array, so instead of using multiple variables to store single TRUE / FALSE values, you can use a single unsigned char variable to store 8 TRUE / FALSE values.
Note: As i have some time out of the C/C++ world, i'm not 100% sure that it's "lives >> 7", but it's with the '>' symbol, a little research on it and you'll be ready to go.

Answer 4

You're correct that a char is the smallest type - and it is typically (8) bits, though this is a minimum requirement. And sizeof(char) or (unsigned char) is (1). So, consider using an (unsigned) char to represent (8) columns.
How many char's are required per row? It's (cols / 8), but we have to round up for an integer value:
int byte_cols = (cols + 7) / 8;
or:
int byte_cols = (cols + 7) >> 3;
which you may wish to store with in the net_t data structure. Then:
calloc(n->rows * n->byte_cols, 1) is sufficient for a contiguous bit vector.
Address columns and rows by x and y respectively. Setting (x, y) (relative to 0) :
n->vec[y * byte_cols + (x >> 3)] |= (1 << (x & 0x7));
Clearing:
n->vec[y * byte_cols + (x >> 3)] &= ~(1 << (x & 0x7));
Searching:
if (n->vec[y * byte_cols + (x >> 3)] & (1 << (x & 0x7)))
/* ... (x, y) is set... */
else
/* ... (x, y) is clear... */
These are bit manipulation operations. And it's fundamentally important to learn how (and why) this works. Google the term for more resources. This uses an eighth of the memory of a char per cell, so I certainly wouldn't consider it premature optimization.