Convert double to uint8_t*
<p>I have a function that accepts uint8_t* which is supposed to be a string. 
I want to send a double to this function. 
I have tried this code below but it doesn't work.</p>

<pre><code>double a = 10.98     
uint8_t* p = (uint8_t*) &amp;a;
printf("p: %u \n", p); 
send_data(p);
</code></pre>

<p>But This code below works, all I want is to replace the string "90" with a double variable ABOVE.</p>

<pre><code>static const char *data[6];
data[0] = "90";
static uint8_t *test; 
test = ( unsigned char *) data[dataCounter] ;
send_data(test);
</code></pre>

<p>So what I mean by doesn't work is that the function send_data is supposed to send a string over bluetooth to a android phone. If I do it like the first sample code, the string is passed correctly. </p>

<p>Note: <strong>I think its possibly because of the difference in data types that is being passed to the second argument. The function is expecting 3 arguments.</strong> </p>

<pre><code> static uint32_t send_data(uint8_t data[]){ 
     return ble_nus_string_send(&amp;m_nus, data, 5);
 }
</code></pre>

<p><strong>This is the function defintion:</strong></p>

<pre><code>uint32_t ble_nus_string_send    (ble_nus_t * p_nus,uint8_t * p_string,
uint16_t    length 
)
</code></pre>