Convert char* to uint8_t

Question

I transfer message trough a CAN protocol.

To do so, the CAN message needs data of uint8_t type. So I need to convert my char* to ui

Answer 1

Is your string an integer? E.g. char* bufferSlidePressure = "123";?

If so, I would simply do:

uint8_t slidePressure = (uint8_t)atoi(bufferSlidePressure);

Or, if you need to put it in an array:

slidePressure[0] = (uint8_t)atoi(bufferSlidePressure);

Edit: Following your comment, if your data could be anything, I guess you would have to copy it into the buffer of the new data type. E.g. something like:

/* in case you'd expect a float*/
float slidePressure;
memcpy(&slidePressure, bufferSlidePressure, sizeof(float));

/* in case you'd expect a bool*/
bool isSlidePressure;
memcpy(&isSlidePressure, bufferSlidePressure, sizeof(bool));

/*same thing for uint8_t, etc */

/* in case you'd expect char buffer, just a byte to byte copy */
char * slidePressure = new char[ size ]; // or a stack buffer 
memcpy(slidePressure, (const char*)bufferSlidePressure, size ); // no sizeof, since sizeof(char)=1