arrays - Conversion between uint8 and char in C -

i have api implements writing operation eeprom. here declaration:

cyble_api_result_t cyble_storeappdata (uint8 * srcbuff, const uint8 destaddr[], uint32 bufflen, uint8 isforcewrite); 

it working when call function , send array parameter srcbuff has been declared uint8 type.

the problem is, need send char array pointer it. thinking char uint8, compiler warning if send char array pointer function instead of uint8. why can't use char instead of uint8 ? here 2 examples of calling function:

static const uint8      datastack_rom[dedicatedromsize] = {0}; uint8                   container_id[10]; char                    prefix[10];  //call function container_id has been declared uint8. working. cyble_storeappdata(container_id,datastack_rom,10,0);  //call function prefix has been declared char. not working. cyble_storeappdata(prefix,datastack_rom,10,0); 

here warning second call:

passing char[10] parameter of type 'uint8 *' converts between pointers integer types different sign.

aren't char , uint8 same?

both types 8bits long. difference comes signedness.

• the uint8 type unsigned.
• the char type should signed in case. actually, compiler dependent, compilers consider char type signed default , have option force char type unsigned if needed. see c99 standard document reference §6.2.5p15:

the implementation shall define char have same range, representation, , behavior either signed char or unsigned char.

char_min, defined in limits.h, have 1 of values 0 or schar_min, , can used distinguish 2 options.