Convert array values of 1's and 0's to binary

Question

In Arduino IDE, I am placing all of input values to an array like so:

int eOb1 = digitalRead(PrOb1);
int eLoop = digitalRead(PrLoop);
int eOb2 = digitalRead(PrOb2);

InputValues[0] = eOb1;
InputValues[1] = eLoop;
InputValues[2] = eOb2;
InputValues[3] = 0;
InputValues[4] = 0;
InputValues[5] = 0;
InputValues[6] = 0;
InputValues[7] = 0;

I would like to convert it to a byte array like so: 00000111.
Can you show me please. I tried using a for Loop to iterate through the values but it doesn't work.

char bin[8];
for(int i = 0; i < 8; ++i) {
   bin &= InputValues[i];
}

What do you mean by byte array? I got confused when I started to answer. You want to get a byte having bits set as per `InputValues`? — user2736738, Feb 28 '18 at 06:31
I want to Serial.println the result as "0000111". The InputValues[0] to InputValues[2] can be 1 or 0 depending on the toggled switch. — stack questions, Feb 28 '18 at 06:32
Please see my updated code that incorporated your previous suggestion. — stack questions, Feb 28 '18 at 06:34
@stackquestions please do not change the question to incorporate a presented solution. It leaves the questions and answers in an inconsistent state. I am rolling back your edit for this one. — Ajay Brahmakshatriya, Feb 28 '18 at 06:39
@stackquestions your last comment is very confusing. Do you want a string or a byte value. They are not *like* each other at all. — Ajay Brahmakshatriya, Feb 28 '18 at 06:41
Am sorry for the confusion. I just want it to be one string of "0000101". Thank you. — stack questions, Feb 28 '18 at 06:52

score 5 · Accepted Answer · answered Feb 28 '18 at 06:35

5

If I understand your requirement correctly, you have an array of individual bits and you need to convert it into a byte that has the corresponding bits.

So to start, you should declare bin to be of type unsigned char instead of char[8]. char[8] means an array of 8 bytes, whereas you only need a single byte.

Then you need to initialize it to 0. (This is important since |= needs the variable to have some defined value).

unsigned char bin;

Now, unsigned char is guaranteed to have 1 byte but not 8 bits. So you should use something like uint8_t IF it is available.

Finally you can set the appropriate bits in bin as -

for(int i = 0; i < 8; ++i) {
    bin |= (InputValues[i] << i);
}

There are two things I have changed.

I used |= instead of &=. This is the bitwise OR operator. You need to use OR because it only sets the correct bits in the LHS and leaves other bits untouched. An AND won't necessarily set that bit and will also mask away (set to 0), the other bits.
Shifted the bit in the array to the corresponding position using << i.

answered Feb 28 '18 at 06:35

Ajay Brahmakshatriya

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It worked! But it is giving me 1 or 2 or 7 or 3. It's giving out hex. Can I have it as byte only like "0000101"? – stack questions Feb 28 '18 at 06:50
`uint8_t` is always available in standard C, unless someone gets the not-too-brilliant idea to write code for various obsolete, oddball TI DSPs in C. – Lundin Feb 28 '18 at 07:52
“have 1 byte but not 8 bits.” really?? Can you show me an example that unsigned char is 1 byte ,but not 8 bits? – peter__barnes Feb 28 '18 at 08:30
@peter__barnes take a look at [this question](https://stackoverflow.com/questions/5516044/system-where-1-byte-8-bit) – Ajay Brahmakshatriya Feb 28 '18 at 09:50

Convert array values of 1's and 0's to binary

1 Answers1