Unsigned int64 to unsigned int32

Fastest and efficient way to convert (unsigned) _uint64 to two (unsigned) _uint32


somefunc(in _uint64 , out _uint32, out _uint32)
Rahul GuptaAsked:
Who is Participating?
developmentguruConnect With a Mentor PresidentCommented:
I use a bit of Delphi magic and do it like this:

procedure TfMain.FormDblClick(Sender: TObject);
  TDDWord = record
    Low, High : DWord;

  UI64 : UInt64;
  DDWord : TDDWord absolute UI64;

  UI64 := 1;
  //DDWord.Low = 1
  //DDWord.High = 0

Open in new window

By simply assigning the UI64 variable, the breakout of the two 32 bit pieces is done, with no extra code, no extra time.  The absolute keyword is for instances just like this, where you want to simply view one area of memory in a different way.  If you run this code in Delphi (with debug info on and optimizations off) then you will be able to inspect that the low value is 1 and the high value is 0, after you do the assignment.  The splitting litterally take no time...
Kent OlsenConnect With a Mentor Data Warehouse Architect / DBACommented:

The fastest way is to just let C do all of the work.

 _uint64 I64;
 _uint32 L32;
 _uint32 U32;

  L32 = I64;
  U32 = (I64 >> 32);

Depending on your compiler options you may wish to recast these items:

  L32 = (_uint32)I64;
  U32 = (_uint32)(I64 >> 32);

Good Luck,
Rahul GuptaAuthor Commented:
does it support both little endian and big endian systems
Free Tool: ZipGrep

ZipGrep is a utility that can list and search zip (.war, .ear, .jar, etc) archives for text patterns, without the need to extract the archive's contents.

One of a set of tools we're offering as a way to say thank you for being a part of the community.

Kent OlsenData Warehouse Architect / DBACommented:
Yep.  The C implementation is specifically mindful of the underlying hardware requirements and all integer type operations MUST work correctly according to the rules.

Now if you were to store the 64-bit value and try to extract it from the memory address and memory address+4 locations, all bets are off.  In this case you'd be at the mercy of the big/little endian rules.

But with integer operations the value is already converted to its "native" value in one of the hardware registers so it's simply a matter of selecting the upper or lower 32 bits.

Rahul GuptaAuthor Commented:
I had done in this way...

   _uint64= _uint64(_uint64(_uint32var1) << 32) | _uint32var2;

  _uint32var1 = _uint32(_uint64variable & &HFFFFFFFF);
  _uint32var2 = _uint32(_uint64variable >> 32);

Open in new window

Kent OlsenData Warehouse Architect / DBACommented:
That's essentially the same.  The only significant different is that on line 3, you mask off the lower 32 bits of the 64 bit value before storing it in a 32-bit container.  The mask is unnecessary since the container can't possibly hold more than 32 bits.

If you need that solution to be mindful of different hardware then you would change the type with conditional compilation and it would be just as fast...
In C/C++ you can get the same effect by having a struct for the two smaller pieces.

struct Two32s {
  _uint32 Low;
  _uint32 High;

union 64BitSplit {
  _uint64 Whole;
  Two32s Split;

I haven't done C/C++ in a long time, so the code is "as is"... just to get the gyst of it.  If someone else feels the need to correct, by all means do!

If you assign to a variable of type 64BitSplit as

MyVar.Whole = VALUE;
You can then read out MyValue.Low and MyValue.High.  The Union makes the two structures occupy the same memory.

Once again, if the CPU independence is important then you should use conditional defines to make the struct swap low for high as needed.
Kent OlsenData Warehouse Architect / DBACommented:
That code will sometimes work, but it's hardware dependent.

It doesn't work for both big and little endian systems.  It should also have the appropriate #PRAGMA or command line option to force data alignment to the 32-bit boundary (or less).

Since you need to compile for little endian versus big endian, conditional compilation takes care of that... I thought I mentioned that already...
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.