Writing over bit boundaries using unions - c++

Question

I am essentially trying to use a union to cast a structure of data pieces with varying bit-widths into a nice clean array of integers. I have written a small program to illustrate my issue.
#include <stdio.h>
#include <iostream.h>
union {
struct {
long blah1;
short blah2;
long blah3;
short blah4;
int blah5;
} data;
int buffer[6];
} db;
int main(int argc, char* argv)
{
db.data.blah1 = 0x1111111111111111;
db.data.blah2 = 0x2222;
db.data.blah3 = 0x3333333333333333;
db.data.blah4 = 0x4444;
db.data.blah5 = 0x55555555;
for(int i=0;i<6;i++) cout << "Word " << i << ": " << std::hex << db.buffer[i] << endl;
}
Output:
Word 0: 11111111
Word 1: 11111111
Word 2: 2222
Word 3: 0
Word 4: 33333333
Word 5: 33333333
Expected Output:
Word 0: 11111111
Word 1: 11111111
Word 2: 33332222
Word 3: 33333333
Word 4: 44443333
Word 5: 55555555
I compiled using gcc version 4.1.2 20080704 (Red Hat 4.1.2-54)
Do I have something formatted incorrectly or am I trying to use this functionality for something other than it was intended? Is there another way to achieve my expected output without having to use bit-wise manipulation and endless shifting?

Answer 1

As pointed out by #happydave and Floris, result could be achieved by using pragma pack with value 1 to stop padding introduced for bit alignment.
#include <stdio.h>
#include <iostream>
using namespace std;
#pragma pack(push,1)
union {
struct {
long blah1;
short blah2;
long blah3;
short blah4;
int blah5;
} data;
int buffer[6];
} db;
#pragma pack(pop)
int main(int argc, char** argv)
{
db.data.blah1 = 0x1111111111111111;
db.data.blah2 = 0x2222;
db.data.blah3 = 0x3333333333333333;
db.data.blah4 = 0x4444;
db.data.blah5 = 0x55555555;
for(int i=0;i<6;i++) cout << "Word " << i << ": " << std::hex << db.buffer[i] << endl;
}