0

Hey I wrote this function to convert an bool array to int: (Math Lob included):

bool b_array[5];

int convertToInt(start_index,stop_index){
    int run=0;
    int result=0;
    for(int id=start_index;id<=stop_index;id++){
        result+=b_array[id] * pow(2.0,run);
        run++;
    }
    return result;
}

but it doesnt this work?

Thanks and kind regards Nic

Improve this question
4
  • "Does not work" is a poor problem description. Explain what it actually does and what you expected it to do. Provide a minimal but complete test sketch, that shows the problem Commented Nov 20, 2019 at 20:51
  • You seem to want to convert an array of bools into the respective integer value, so e.g. { false, true, true, false} into 0b0110 = 6? Then you should use better casting, so (int)b_array[id] and (int)pow(2.0,run). Though it's a weird way to do it, since bitshifting would be more appropriate (result += b_array[id] ? (1 << run) : 0) . Also be aware of the numerical limits of your 16-bit signed int. Commented Nov 20, 2019 at 22:16
  • The pow function returns a float. Floats are inherently inaccurate. The index to the array has to be an int. When you do powers of 2 with pow you don’t actually get 4. You get 3.9999999 something. So when that gets truncated it turns into 3 and not the 4 that you want. Use bit shifting for integer powers of 2 and you won’t have this problem. Commented Nov 21, 2019 at 15:45
  • Thank you Guys a lot! It worked like a charm but I could not have made it without you! Commented Nov 22, 2019 at 12:43

1 Answer 1

Reset to default
2

Assuming little-endian bools (least significant bool first), you should use simple bit-shifting, not floating point power calculations:

for(int id=start_index;id<=stop_index;id++){
    result |= b_array[id] ? (1 << run) : 0;
    run++;
}

The ternary operation (b_array[id] ? (1 << run) : 0) means:

  • If b_array[id] is true, then
    • Or result with 1 left-shifted run times and store in result
  • otherwise
    • Or result with 0 and store the result (no-operation).
Improve this answer
2
  • You can simplify this in C++ because bool value promoted to an integer is 0 or 1. So (b_array[id] << run) can be used. In C there is no bool data type, so something like ((b_array[id] !=0) << run) should be used. And i like for example result = (result<<1) | b_array[id]; more (it can be result = result*2 + b_array[id] too) Commented Nov 22, 2019 at 9:05
  • Thank you for your Solution and your explaination, it worked on first try like a charm! Thank you a lot, I could not have solved that without your help! Commented Nov 22, 2019 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.