Convert array of bits into a hexadecimal number

Question

What would be the best way to go above converting from an integer array representing the bits of a number to hexadecimal?

My current attempt reads four bits at a time then tries to print the corresponding hex character, but I dont see to get any output.

Here is what I have:

 /**
  * Maps a 4 bit string in binary to the corresponding
  * hex digit
  */
 void map_bin_to_hex(char *c)
   {
      printf("In mapbin: C = %s\n", c); //This is printing funny output
      if(!strcmp(c, "0000"))      printf("0");
      else if(!strcmp(c, "0001")) printf("1");
      else if(!strcmp(c, "0010")) printf("2");
      else if(!strcmp(c, "0011")) printf("3");
      else if(!strcmp(c, "0100")) printf("4");
      else if(!strcmp(c, "0101")) printf("5");
      else if(!strcmp(c, "0110")) printf("6");
      else if(!strcmp(c, "0111")) printf("7");
      else if(!strcmp(c, "1000")) printf("8");
      else if(!strcmp(c, "1001")) printf("9");
      else if(!strcmp(c, "1010")) printf("A");
      else if(!strcmp(c, "1011")) printf("B");
      else if(!strcmp(c, "1100")) printf("C");
      else if(!strcmp(c, "1101")) printf("D");
      else if(!strcmp(c, "1110")) printf("E");
      else if(!strcmp(c, "1111")) printf("F");
   }

 /**
  * Reads 4 array elements at a time, passing them to map_bin_to_hex
  */
 void bin_array_to_hex(int *b, int length)
 {
    int i, j, k;

    //read 4 bits 16 times
      for(i = 0; i < 64; i+=4)
      {
         char hexB[4];
         for(k = 0, j = i; j < (i+4); j++, k++)
         {
            hexB[k] = b[j];
            printf("h[%d] = %d\n", k, hexB[k]);
         }
         map_bin_to_hex(hexB);
      }
      printf("\n");
  }

 int main()
 {
   // a 64-bit number represented as an array of bits
   int x[] =[1,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1,1,0,0,1,1,1,1,1,1,1,1,
             1,1,1,1,0,0,0,0,1,0,1,0,1,0,1,0,1,1,1,1,0,0,0,0,1,0,1,0,0,0,1,0];
   bin_array_to_hex(x, 64);
  }

The bin_array_to_hex() function reads the array in 4bit sections then passes those to map_bin_to_hex(), the output from bin_array_to_hex() is is correct, but something is wrong with my mapping function.

Any help would be appreciated.

EDIT: New Solution

   void map_bin_to_hex(char *c)
   {
        char hexMap[] = 
        {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};

        int index = strtoull(c, NULL, 2);
        printf("%c", hexMap[index]);
   }

Converting to char is a bit heavy. I'd be using bit shifting to get an integer and then sprintf — David Heffernan
– David Heffernan, Commented Feb 12, 2012 at 20:40
I couldn't seem to think of a better solution, it there an easier way to convert from an array of bits to a hex value? — Hunter McMillen
– Hunter McMillen, Commented Feb 12, 2012 at 20:42

Community · Accepted Answer · 2017-05-23 12:33:41Z

While your program will run with dasblinkenlight's answer, it could run much faster if you wouldn't use strcmp. First of all, you could write yourself a function that prints the hex of a byte:

void printHexChar(unsigned char value){
    unsigned char lower = value & 0xf; // Use binary 'and' to mask the lower byte
    unsigned char upper = (value & 0xf0) >> 4; // Use binary 'and' to mask upper byte

    if(lower >= 10) // If lower is in range [10-15], than add a value [0-5] on 'a'.
        lower = 'a' + (lower - 10);
    else
        lower = lower + '0'; // It's in range [0-9], so we have to add it to '0'.

    if(upper >= 10) // Same as lower
        upper = 'a' + (upper - 10);
    else
        upper = upper + '0';

    printf("%c%c",upper,lower); // Print out the hexadecimal number
}

This will enable you to print uint8_t (unsigned integers with a length of 8 bits) in hexadecimal. If you don't know the >> operator: it shifts all bits n times right. For example 23 >> 2 = 5 since 23 = [10111], [10111] >> 2 = [101] = 5. The operator << will do the opposite and shift the bits left.

void bin_array_to_hex(int *b, int length){
    if(length % 8){
        printf("Must be dividable by eight!\n");
        return;
    }
    unsigned int i;
    unsigned int j;
    for(i = 0; i < length; i = j){
        unsigned char a = 0; // Has length of 8 bits
        for(j = i; j < i+8; ++j){ // take 8 bits...
            a |= b[j]<<(j-i); // and set them in a
        }
        printHexChar(a);
    }           
}

 int main(){
   // a 64-bit number represented as an array of bits
   int x[] ={1,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1,1,0,0,1,1,1,1,1,1,1,1,
             1,1,1,1,0,0,0,0,1,0,1,0,1,0,1,0,1,1,1,1,0,0,0,0,1,0,1,0,0,0,1,0};
   bin_array_to_hex(x, 64);
  }

PSN · Accepted Answer · 2017-02-11 17:18:58Z

2

You are almost there,

replace

hexB[k] = b[j];

with

hexB[k] = b[j] + '0';

Also, char hexB[4]; needs to be char hexB[5]; (last char is for null termination; don't forget to set hexB[4]='\0'; before calling map_bin_to_hex).

Once you're done with this initial fix, take a look at other solutions available on the internet: you will see how to do the whole thing in five or six lines.

EDIT As an additional challenge, see if you can use these two string constants to avoid the long chain of if-then-else statements in your first method:

"0000000100100011010001010110011110001001101010111100110111101111"
"0123456789abcdef"

Hint: observe how the first string is exactly four times longer than the second one.

edited Feb 11, 2017 at 17:18

PSN

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answered Feb 12, 2012 at 20:34

Sergey Kalinichenko

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5 Comments

Hunter McMillen Over a year ago

Thanks, this worked perfectly. Would you mind telling me what adding '0' to b[j] is doing? Is that some sort of implicit conversion?

Dervall Over a year ago

You are printing character number 1 in your printf. Adding '0' offsets it to make it into either the character '0' or '1'

Sergey Kalinichenko Over a year ago

@HunterMcMillen Dervall explained it exactly right: your strcmp expects strings of ASCII digits, but your original program was passing strings of non-printable characters.

Hunter McMillen Over a year ago

@dasblinkenlight Probably not what you had in mind with your clue, but I have a new solution that is much shorter, I posted it above.

Sergey Kalinichenko Over a year ago

@HunterMcMillen That's very good. You could shorten it some more by replacing the array of characters initializer with a string constant #2 from my edit - it is the same thing: char hexMap[] = "0123456789ABCDEF";

Jor-el · Accepted Answer · 2013-02-21 19:25:10Z

0

For the answer by Zeta, the case of =10, should be included in the first conditions itself. Thus, the modified code will look like

if(lower>=10)

answered Feb 21, 2013 at 19:25

Jor-el

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