Converting 8 bits in binary to 1 byte

Question

I have a string of 8 bits and I want to convert it into 1 byte. I am not sure why my function is not working properly. I have 8 bits stored into an array of 8 unsigned chars. This is my method so far:

unsigned int bitsToBytes(unsigned char *bits)
{
  unsigned int sum = 0;
  for(int i = 8; i >= 0; i--)
  {
    sum += bits[i];
    sum<<=1;
  }
  return sum;

}

int main()
{
  unsigned char bits[8]='01010111';
  unsigned int byt;
  byt = bitsToBytes(bits);
  cout << byt; //doesn't give me the right result
}

Could anyone help to get correct result? Thanks alot!

With your terminology being completely off, it's hard to know what it is you really want done. What's the "right result" that you didn't get? — sweenish, Jul 20 '20 at 20:08
Among all the other errors you have an out of bounds array access at the start of your loop. `8` is not a valid index for an array of size 8. Array indexes start at zero. Try `for(int i = 7; i >= 0; i--)` — john, Jul 20 '20 at 20:11
@LiamLony: you can accept one of the answers by clicking on the grey checkmark below its score. — chqrlie, Jul 25 '20 at 12:18

chqrlie · Answer 1 · 2020-07-20T21:53:30.687

There are multiple problems:

the loop in your code iterates 9 times, from 8 to 0 inclusive,
you read bits[8] which is an invalid access.
you add bits[i] which is a character with value '0' or '1', not a bit value.
the bits are probably shifted in the wrong order.
in main(), the initializer for the array is incorrect: unsigned char bits[8] = '01010111'; should be unsigned char bits[8] = "01010111";

Here is a modified version:

unsigned int bitsToBytes(unsigned char *bits) {
    unsigned int sum = 0;
    for (int i = 0; i < 8; i++) {
        sum = (sum << 1) + (bits[i] - '0');
    }
    return sum;
}

int main() {
    unsigned char bits[8] = "01010111";
    unsigned int byt;
    byt = bitsToBytes(bits);
    cout << byt;
    return 0;
}

0___________ · Answer 2 · 2020-07-20T20:29:41.430

2

Issues:

bits is to short to accommodate 8 chars + null char
shuft of the sum has to be done before adding the value
Strings are enclosed in " not '

unsigned int bitsToBytes(unsigned char *bits)
{
  unsigned int sum = 0;
  while(*bits)
  {
    sum<<=1;
    sum += *bits++ - '0';
  }
  return sum;

}


int main()
{
  unsigned char bits[]="0101011101100101010101";
  unsigned int byt;
  byt = bitsToBytes(bits);
  std::cout << (unsigned)byt; //doesn't give me the right result
}

https://godbolt.org/z/PjjbGK

https://godbolt.org/z/njaKPn

edited Jul 20 '20 at 20:29

answered Jul 20 '20 at 20:11

0___________

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Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackoverflow.com/rooms/218248/discussion-on-answer-by-p-j-converting-8-bits-in-binary-to-1-byte). – Samuel Liew Jul 21 '20 at 04:21

Converting 8 bits in binary to 1 byte

2 Answers2