Converting a char to binary bits, i.e.: 'a' to binary equivalent-CodePudding

I'm hoping to get a quick hand with my code. I need to convert char c to binary before my function does other things. However, I am getting errors that I am using itoa wrong and am not sure how else to even attempt this part of the function.

The main function is:

char *alpha = "abcde";
unsigned char bits[8];

for (int pos=0;pos<5;pos  )
{
    printf( "%c ", alpha[pos] );
    char2bits( alpha[pos], bits );
}

The function, for what I have so far, is:

void char2bits( char c, unsigned char bits[8] ){
    for(int i = 0; i < 5; i  ){
        itoa(c, (bits)[i], 2);
    }

    // ...
}

CodePudding user response：

A possible implementation of char2bits():

void char2bits(char c, unsigned char bits[8]) {
  for (size_t i = 0; i < 8; i  = 1) {
    bits[i] = (c >> (7 - i)) & 1;
  }
}

int main(void) {
  char c = 3;
  unsigned char bits[8];

  char2bits(c, bits);
  // now `bits` is equal to {0, 0, 0, 0, 0, 0, 1, 1}

  for (size_t i = 0; i < 8; i  = 1) {
    printf("%hhu\n", bits[i]);
  }

  return 0;
}

CodePudding user response：

First when you want to work with bytes, you should use unsigned char or uint8_t.

This function change a byte into a printable string (you can use printf or puts with "bits" array):

void char2bits(unsigned char c, unsigned char bits[8]) {
    int shift, i;
    for(shift = 7, i = 0; i < 8; --shift,   i)
        bits[i] = ((c >> shift) & 0x1)   '0';
}

What happend ?

with c = 250 = 1111 1010

first your bits table is empty so: bits = ""

Step i = 0:

(c >> shift) = shift right by 7 = 0000 0001

(c >> shift) & 0x1 = keep only last bit = 0000 0001

((c >> shift) & 0x1) '0' = ((c >> shift) & 0x1) 48 = 48 1 = 49 = '1' in ascii

so bits = "1";

Step i = 1:

(c >> shift) = shift right by 6 = 0000 0011

(c >> shift) & 0x1 = keep only last bit = 0000 0001

((c >> shift) & 0x1) '0' = ((c >> shift) & 0x1) 48 = 48 1 = 49 = '1' in ascii

so bits = "11";

Step i = 2:

(c >> shift) = shift right by 5 = 0000 0111

(c >> shift) & 0x1 = keep only last bit = 0000 0001

((c >> shift) & 0x1) '0' = ((c >> shift) & 0x1) 48 = 48 1 = 49 = '1' in ascii

so bits = "111";

Step i = 3:

(c >> shift) = shift right by 4 = 0000 1111

(c >> shift) & 0x1 = keep only last bit = 0000 0001

((c >> shift) & 0x1) '0' = ((c >> shift) & 0x1) 48 = 48 1 = 49 = '1' in ascii

so bits = "1111";

Step i = 4:

(c >> shift) = shift right by 3 = 0001 1111

(c >> shift) & 0x1 = keep only last bit = 0000 0001

((c >> shift) & 0x1) '0' = ((c >> shift) & 0x1) 48 = 48 1 = 49 = '1' in ascii

so bits = "11111";

Step i = 5:

(c >> shift) = shift right by 2 = 0011 1110

(c >> shift) & 0x1 = keep only last bit = 0000 0000

((c >> shift) & 0x1) '0' = ((c >> shift) & 0x1) 48 = 0 48 = 48 = '0' in ascii

so bits = "111110";

Step i = 6:

(c >> shift) = shift right by 1 = 0111 1101

(c >> shift) & 0x1 = keep only last bit = 0000 0001

((c >> shift) & 0x1) '0' = ((c >> shift) & 0x1) 48 = 1 48 = 49 = '1' in ascii

so bits = "1111101";

Step i = 7:

(c >> shift) = shift right by 0 = 1111 1010

(c >> shift) & 0x1 = keep only last bit = 0000 0000

((c >> shift) & 0x1) '0' = ((c >> shift) & 0x1) 48 = 0 48 = 48 = '0' in ascii

so bits = "11111010";

I hope that will help you.

CodePudding user response：

Here's how I would do it

void char2bits( char c, unsigned char bits[8] )
{
    for( int i = 0; i < 8;   i)
    {
        // create a mask for bit `i`
        int mask = 1 << i;

        // set the corresponding member of the `bits` array to 0 or 1
        // depending on the value of the relevant bit
        bits[i] = (c & mask) ? '1' : '0';
    }
}

Hopefully this is clear.

The basic technique is to use the bitwise and & operator to strip out a single bit from the input character. The mask will have values 1, 2, 4, 8 etc. as the loop executes. I feed the result of the & into the ternary operator test ? a : b to store either a "1" or a "0" character in the output array.

This is equivalent to writing out

if(c & mask)
    bits[i] = '1';
else
    bits[i] = '0';

Note that I've iterated over all 8 possible bits, rather than the 5 that your code used. Note, too, that in the output, bits[0] will hold the least significant bit of the input character c.