RE: Device name has only 4 letters

smackenzie — Thu, 16 Dec 2021 00:44:01 GMT

This:

const uint8_t * p_device_name =&device_namee[8];

Is getting the address of the 8th character in `device_name`, so the p_device_name pointer is pointing to the memory after the string.

You can either change to

char device_namee[9] ="EF234ABC"; 
const uint8_t * p_device_name = device_namee;

(needs to be 9 to accommodate for the null terminator byte)

Or just not bother with the other variable and go with

static const char* device_name ="EF234ABC";

static void gap_params_init(void)
{
  ...
    err_code = sd_ble_gap_device_name_set(&sec_mode,
                                          device_name,
                                          strlen(device_name));
   ...
}

RE: Device name has only 4 letters

Rob Garner — Wed, 15 Dec 2021 23:32:43 GMT

Hi Hani

I noticed that you declare device_namee like this: char device_namee[8] ="EF234ABC".

So basically you are declaring an array with 8 bytes of type char.

Then you set a pointer to this array, as follows: const uint8_t * p_device_name =&device_namee[8]

I believe that you should declare the pointer like this: const char * p_device_name =&device_namee (no array size identifier here!)

That's my two cents. Let me know how that goes.

Cheers

Rob