Question

Basically in the code below, my final array does not seem to have the contents from function1(). Any ideas on why I can’t get this to work ? Thanks.

    #include <stdio.h>
    #include <string.h>
    #include<stdlib.h>

  unsigned char *function1() 
 {
  unsigned char array2[] = { 0x4a,0xb2 };
  return (array2 );

  }

   main()

     {
unsigned char temp[] = { 0xaa, 0x0b, 0x03,0x04,0x05,0x06,0x07,0x08,0x09 };
unsigned char x[2];
unsigned char *packet;
int pkt_len;
pkt_len = sizeof(temp) + sizeof(x);
packet = (unsigned char*) malloc ( pkt_len +1);

memset( packet, 0x00, pkt_len +1);
unsigned char *pointer1 = malloc ( sizeof(temp) + 1);

memset( pointer1, 0x00, sizeof(temp) +1);
memcpy (pointer1, temp, sizeof(temp) );

memcpy (packet, pointer1, sizeof(temp) );
printf("\nPacket before copy is 0x%x\n", packet[8]);

    unsigned char *array2 = malloc ( sizeof (x) + 1)  ;
    array2 = (char *)function1();
printf("\nArray2 is 0x%x\n", array2[0]);
    memcpy (packet + sizeof(temp), array2, sizeof(x) );
printf("After copy, Packet contents are 0x%x\n", packet[9]);
  }

Answer 1

Following are the mistakes that I observed in your code.
You wrote

  char temp [] = { a,s,d,f,g,h};
  char * pointer1, *array1;
  pointer1 = &temp;  
  memcpy (array1, pointer1, sizeof( temp) );

Now there is no need to do this pointer1 = &temp, name of any array itself is a pointer.
Hence you can simply do

  char temp [] = { a,s,d,f,g,h};
  char *pointer1;
  memcpy (pointer1, temp , sizeof( temp) );

But wait!

Does pointer1 has enough space to store the contents of temp[]? In your code you have not assigned any space to pointer1, which is likely to crash your program.

The correct way to do it is

  char temp [] = { 'a','s','d','f','g','h'};
  char *pointer1 = malloc( sizeof(char) * (sizeof( temp) + 1) );
  memset( pointer1, 0x00, sizeof( temp) + 1 );
  memcpy (pointer1, temp , sizeof( temp) );

here before copying any value into pointer1 we made sure that it has got enough space.

No need to cast malloc() retrun value. In sizeof( temp) + 1 1 is added for null character. Then we did memset() which filled the memory pointed to by pointer1 with null. Just good and healthy practice.

Then you

memcpy ( pointer1 + sizeof(temp), pointer3, sizeof ( the temp array)

Again, does pointer1 has enough space for contents of pointer3? Do you own the memory area pointed by pointer1 + sizeof(temp)? It too will crash your program.
Now you either use realloc() or assign a bigger space to pointer1 with malloc() at earlier stage.

Why sizeof ( the temp array) here? Don't you think it should have the number of bytes in pointer3?

Finally in the definition of function1()

char *pointer2, *array2;
 // Now i need to have pointer2 point to contents of array2.
 pointer2 = &temp2;
 return pointer2 

What does array2 do? Nothing! Then it should be removed.
To return just use

return temp2;

which means pointer2 is also useless.

Hope it helps.