Need help with this memcpy and array code

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I want to merge the 2 arrays in the main function. But for some reason, even though I can get array3[0] and array3[1] to have the right values, when they are copied into the packet, their values are different. Any idea why ?
Thanks for your help.

#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]);
}

Posted 25-Jun-10 8:15am

talk2vogon

Updated 25-Jun-10 9:36am

Aescleal

v3

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Aescleal · Accepted Answer · 2010-06-25T10:54:00

First off: Are you sure you want to return an address of a local variable from function1()? There's no guarentee that array2 will point to a meaningful block of memory when the function returns.

Secondly: Are you sure your program is valid C? I'm not an expert on C anymore but declaring variables after non declaring statements was a complete no-no in C89. array2 and pointer1 are the variables I'm looking at in particular.

Thirdly: You've got loads of redundant copying and memsetting going on so it's a bit hard to see what you're actually intending to do. For example:

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

could be replaced with:

memcpy ( pointer1, temp, sizeof temp );
pointer1[ sizeof temp] = 0x00;

Fourthly: In C you don't need to cast the return value of malloc. void * is compatible with any pointer type.

Fifthly: Whenever an experienced C programmer sees sizeof( x ) they start looking for a type called x. Leave off the braces when taking the size of an object.

Sixth: What's the point of pointer1? It seems totally redundant.

If all you want to is dynamically create a lump of memory and copy two other arrays into it you can do it in about three lines:
- allocate the memory
- copy the first array into the allocated memory
- copy the second array into the allocated memory after the first array

Cheers,

Ash