I have this array of bits
int bits[8] = { 0, 1, 1, 0, 0, 1, 0, 1 }
This is 65 in hex or 101 in decimal. The ASCII letter is 'e'. How do I go about reading my array into a char and int (the decimal value)?
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1I'd look at this: stackoverflow.com/questions/13667746/…Michael McPherson– Michael McPherson2018-06-25 14:34:46 +00:00Commented Jun 25, 2018 at 14:34
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2Second half of the accepted answer is wrong though, sadly...Baum mit Augen– Baum mit Augen ♦2018-06-25 14:39:09 +00:00Commented Jun 25, 2018 at 14:39
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Do prefer the second up-voted answer.YSC– YSC2018-06-25 14:42:42 +00:00Commented Jun 25, 2018 at 14:42
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1 Answer
You could use bit shifting in order to get the char from the bit array like so:
int bits[8] = { 0, 1, 1, 0, 0, 1, 0, 1 };
char result = 0; // store the result
for(int i = 0; i < 8; i++){
result += (bits[i] << (7 - i)); // Add the bit shifted value
}
cout << result;
This basically loops through your array, bitshifts by the correct amount, and then adds the value to an aggregating "result" variable. The output should be "e".
3 Comments
UKMonkey
I'd probably throw in an
assert(bits[i] ==0 || bits[i] == 1) just because if it fails, the result will silently return a result.
YSC
And I'd use
std::bitset because C != C++ (except if C is a floating point value large enough).
Dorin Botan
for(int i = 0; i < 8; i++) { result = (result << 1) | bits[i]; } would work too and imho. is slightly easier to understand. (Neither my nor the answer proposed by Keveloper check that the values of bits[] are either 0 or 1). Better do a bits[i] == 0 ? 0 : 1 instead of just bits[i].