Is there a way to set the value of an variable by the bits to comprise it?

Question

I am trying to make a hamming code decoder and encoder in C and I cannot find a way to set the bits of a variable individually.

For example, I am trying to somehow do the following:

#include "stdio.h"

int main () {
    short block = 0010101110001110; // variable to contain the bits to decode
}

Clearly this will not work but I am wondering if there is a way to do this or will I have to define it as the actual number this represents?

Do you know how to perform "bitwise OR" and "left shift (<<)"? — Fe2O3, Feb 17 '23 at 04:29
Yes, but I am not sure how this will help me assign the variable, only update it after it is already set. What are you suggesting? — Nasser Kessas, Feb 17 '23 at 04:31
Yes, this involves "composing" the bit pattern in some register... I can't figure out how you plan to create blocks of Hamming codes with a single assignment... — Fe2O3, Feb 17 '23 at 04:40
That's not exactly what I was trying to do, I was just trying to temporarily set the value of a variable to a pre-encoded hamming block so I can test a decode function. — Nasser Kessas, Feb 19 '23 at 23:34

dbush · Accepted Answer · 2023-02-17T04:22:03.373

1

You can use hexadecimal representation, where each digit represents exactly 4 bits.

unsigned short block = 0x2b8e;

edited Feb 17 '23 at 04:22

answered Feb 17 '23 at 04:21

dbush

205,898
23
218
273

Is there any way to do this bit for bit? – Nasser Kessas Feb 17 '23 at 04:21
2

@NasserKessas If you mean in binary, not portably (i.e. some compilers support a `0b` prefix but not all). Hex is just as good and more readable. – dbush Feb 17 '23 at 04:24

score 1 · Answer 2 · answered Feb 17 '23 at 05:04

The next version of C C2x is expected to support binary constants like 0b0010101110001110.

For now, consider hexadecimal constants or perhaps a macro.

Tip: typically such code works best with unsigned types.

#define BIN16U(a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p) ( \
   ((unsigned)(a)<<15) | ((b)<<14) | ((c)<<13) | ((d)<<12) | \
             ((e)<<11) | ((f)<<10) | ((g)<< 9) | ((h)<< 8) | \
             ((i)<< 7) | ((j)<< 6) | ((k)<< 5) | ((l)<< 4) | \
             ((m)<< 3) | ((n)<< 2) | ((o)<< 1) | ((p)<< 0))

unsigned short block = BIN16U(0,0,1,0, 1,0,1,1, 1,0,0,0, 1,1,1,0);

Nasser Kessas, We could use macro magic to take `M(0010101110001110)` into a hex or octal constant and then extract the bits to form the 16-bit value, but that cleverness is useful only in narrow cases. — chux - Reinstate Monica, Feb 17 '23 at 05:13

Is there a way to set the value of an variable by the bits to comprise it?

2 Answers2