Array of 1's and 0's to hex

Question

I have an array of 1's and 0's in C.

int foo[8];

int main()
{
  int i = 0;
  for (i; i < 8; i++)
  {
      foo[i] = 0;
  }

 foo[1] = 1;
 foo[3] = 1;
}

So, my array now is something like this:

01010000

And I want to convert this bin representation of my array to hex number. In this case it will be 0x50.

Is there any fast and easy way to do this?

Group 4 elements, compute the decimal value and look the corresponding digit in a lookup table. — Mihai Maruseac, May 14 '14 at 21:42
A string in C is represented as an array of characters terminated with a NULL byte. A string **is** an array. — aglasser, May 14 '14 at 21:49
@aglasser So, it is a good idea to convert int array to char array? — JohnDow, May 14 '14 at 21:50
Converting from int to char is trivial, but I don't see how it is any different here. You can easily adapt the code given for an int array or a char array. Modify the `do...while` to another for loop that just increments i instead of incrementing the a pointer. int b = *a=='1'?1:0 is just assigning an int value to b whether *a (the current char) is a 1 or 0. Trivial to implement with ints. [Post in question is here.](http://stackoverflow.com/a/5307692/2599996) — aglasser, May 14 '14 at 21:54
I've rolled back your edit. Altering your question to add the solution is not how StackOverflow works. Please see [Can I answer my own question?](http://stackoverflow.com/help/self-answer) for the proper way to do so. Thanks. — Ken White, May 14 '14 at 22:42
What do you mean a hex number? Print it out in hex, or a string of hex rep? — SwiftMango, May 14 '14 at 22:56
I think I'd be using `int value = 0; for (i = 0; i < 8; i++) value = (value << 1) | foo[i];` to get the integer corresponding to the binary digits. This can then be formatted to `0x50` with `char hex[10]; snprintf(hex, sizeof(hex), "0x%.2X", value);` (or just use `printf("0x%.2X\n", value);`, of course). — Jonathan Leffler, May 14 '14 at 22:57

score 0 · Answer 1 · answered May 14 '14 at 22:54

Something like this, for some adaptability?

int start = 0, size = 4, i, value = 0;
for (i=start;i<size+start;i++) {
    value *= 2;
    if (foo[i] == '1') {
        value ++;
    }
}

Simple modular arithmetic. value is going to end up as 0-15, which you can easily convert to a hexadecimal character, and iterate the loop to collect as many hex digits as needed/available.

Array of 1's and 0's to hex

1 Answers1