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

uint8_t is 8 bits of memory, and can store values from 0 to 255

char is probably 8 bits of memory

char * is probably 32 or 64 bits of memory containing the address of a different place in memory in which there is a char

First, make sure you don't try to put the memory address (the char *) into the uint8 - put what it points to in:

char from;
char * pfrom = &from;
uint8_t to;
to = *pfrom;

Then work out what you are really trying to do ... because this isn't quite making sense. For example, a float is probably 32 or 64 bits of memory. If you think there is a float somewhere in your char * data you have a lot of explaining to do before we can help :/