c++ - 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 uint8_t. With my research on this site, I produce this code :

    char* bufferSlidePressure = ui->canDataModifiableTableWidget->item(6,3)->text().toUtf8().data();//My char*

    /* Conversion */
    uint8_t slidePressure [8];
    sscanf(bufferSlidePressure,"%c",
        &slidePressure[0]);

As you may see, my char* must fit in sliderPressure[0].

My problem is that even if I have no error during compilation, the data in slidePressure are totally incorrect. Indeed, I test it with a char* = 0 and I 've got unknow characters ... So I think the problem must come from conversion.

My datas can be Bool, Uchar, Ushort and float.

Thanks for your help.

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