c - Convert double to uint8_t* -

i have function accepts uint8_t* supposed string. want send double function. have tried code below doesn't work.

double = 10.98      uint8_t* p = (uint8_t*) &a; printf("p: %u \n", p);  send_data(p); 

but code below works, want replace string "90" double variable above.

static const char *data[6]; data[0] = "90"; static uint8_t *test;  test = ( unsigned char *) data[datacounter] ; send_data(test); 

so mean doesn't work function send_data supposed send string on bluetooth android phone. if first sample code, string passed correctly.

note: i think possibly because of difference in data types being passed second argument. function expecting 3 arguments.

 static uint32_t send_data(uint8_t data[]){       return ble_nus_string_send(&m_nus, data, 5);  } 

this function defintion:

uint32_t ble_nus_string_send    (ble_nus_t * p_nus,uint8_t * p_string, uint16_t    length  ) 

there 2 different things might mean "sending double string". might mean "send actual existing bytes of double array of bytes" (in other words, send 4.5 {0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x12, 0x40}), or might mean "send textual representation of double string (iow, send 4.5 "4.5"). former case be:

double d = 4.5; ble_nus_string_send(&m_nus, (uint8_t *)(&d), 8); 

this not want, since app sending must expecting same way, same endianness, same floating point representation, etc. want second case:

double d = 4.5; char temp[20]; snprintf(temp, 20, "%g", d); ble_nus_string_send(&m_nus, (uint8_t *)temp, strlen(temp));