Packing 64 bit pointer into 4 WORD's - c++

Question

I need to pack this pointer (which is 64 bits wide) into 4 WORD's, and then later in some other part of the code I need to extract (assemble) these words back into this pointer.
the code looks like this:
std::vector<WORD> vec;
vec.push_back( this ); // how?
later in the code;
pThis = vec.at(0); // how?
I did take a look at LOWORD/HIWORD and LOWBYTE/HIBYTE macros however I still have no idea how would I go about this.
If you ask why on earth would anyone need this, here is why:
I need to fill in creation data of DLGITEMTEMPLATEEX structure which takes WORD as last argument to specify size, data following this is where you put your data, my data is this pointer, and since I'm working on words (std::vector<WORD>) to fill the structure, the last data is 4 WORDS (64 bits) representing the pointer!
Any suggestion or sample code is welcome.

Answer 1

Well, the best way would be to define a struct that derive form DLGITEMTEMPLATEEX. That way, you can avoid to manually do the conversion. While it might not be defined from the standard, it will works on Windows. That kind of code if platform specific anyway!
struct MyTemplate : DLGITEMTEMPLATEEX
{
MyTemplate(MyPointerType *myVariable)
: extraCount(sizeof *this - sizeof(DLGITEMTEMPLATEEX))
, myVariable(myVariable)
{ }
MyPointerType *myVariable; // 64 if compiled for 64 bit target.
};
And when using the data, you do a static_cast to convert back to that structure.
You could add a few static_assert and assertion to validate that it works as intended.

Answer 2

You could use simple bitshifting to seperate the 64bit into 16bits.
#include <stdio.h>
int main(){
uint64_t bigword = 0xDEADBEEFADEADBED;
uint16_t fourth = bigword ;
uint16_t third = bigword >> 16;
uint16_t second = bigword >> 32;
uint16_t first = bigword >> 48;
printf("%llx %x %x %x %x\n",bigword,fourth,third,second,first);
return 0;
}
Then reverse the process when shifting the words back into the 64bit.