uint8t

I'm trying to "translate" an array of uint8_t [uint8_t lets_try[16]] to a string of 16*8+1[null character] elements. For example: lets_try[0] = 10101010 lets_try[1] = 01010101 ... and I would like to have a string like: 1010101001010101...[\0] Here the questions: 1) is there a quick way to perform this operation? I was trying to do it on my own; my idea was starting from translating a single uint8_t variable into a string and obtaining the full array with a loop [I haven't done this last part yet]. At the end I wrote this code: int main() { uint8_t example = 0x14; uint8_t *pointer; char *final

I wrote the following code: #include <iostream> #include <iomanip> #include <stdint.h> using namespace std; int main() { uint8_t c; cin >> hex >> c; cout << dec << c; return 0; } But when I input c —hex for 12—the output is also c . I was expecting 12. Later I learned that: uint8_t is usually a typedef for unsigned char . So it's actually reading c as ASCII 0x63. Is there a 1 byte integer which behaves as an integer while doing I/O and not as char? Not that I know of. You could do the I/O using a wider integer type, and use range checking and casting as appropriate. I'm afraid I don't know of

I have an uint8_t and I need to read/write to specific bits. How would I go about doing this. Specifically what I mean is that I need to write and then later read the first 7 bits for one value and the last bit for another value. edit: forgot to specify, I will be setting these as big endian You're looking for bitmasking . Learning how to use C's bitwise operators: ~ , | , & , ^ and so on will be of huge help, and I recommend you look them up. Otherwise -- want to read off the least significant bit? uint8_t i = 0x03; uint8_t j = i & 1; // j is now 1, since i is odd (LSB set) and set it? uint8