Help, newbie question. Copy Long into an unsigned char array issue

Hello,

I am a C programming newbie. This is my first program and I have made it through quite well, researching tutorials on the web etc. I have a hurdle I just cannot figure out;

Code:

---

int main()
{
 /* Ask for a position input */
 bool ValueOk = false;
 unsigned char Posn[4] = {0};
 unsigned char Temp[1] = {0};
 long double a;
 long b;
 int i;
 int ulIdx;

 do
 {
  cout << "Enter Desired Position (0.00 -> 195.00 mm):  ";
  scanf("%Lf", &a);
  if (a >=0.00 && a<= 195.00)
  {
  if (a < 0)
  {
    b = floor(a * 100);
  } else
  {
    b = ceil(a * 100);
  }
  printf("%02X", b);
  Posn[0] = b;        <---- PROBLEM
  printf("%02X", Posn[0]);
  /* Invert Order Of Bytes For abSendData */
  for (ulIdx = 0; ulIdx < 4; ++ulIdx)
  {
    Temp[ulIdx + (3 - ulIdx)] = Posn[ulIdx];
  }
  for (i=0; i<4; ++i)
  {
    Posn[i] = Temp[i];
  }
  ValueOk = true;
  }
 } while (!ValueOk);
}



---

How do I copy the variable "b" (long), represented as (4) byte values, into the Posn[0, 1, 2, 3] array? I also, want to invert the order, hence the reason for the loop above. I want to then shuffle the (4) bytes in the Posn array, as Posn[3, 2, 1, 0]. How is this done? I've been playing with sprintf, without success. Would someone be kind enough to explain this to me?

Many thanks!

You'll need to mask the number with a bitwise and(&) and shift it down a number of bits to store it in a 1 byte context.

A couple examples:

Code:

---

posn[0] = (0xff000000 & b) >> 24;
posn[1] = (0x00ff0000 & b) >> 16;

---

I can see what the mask is doing, but what does the " >> 24" and " >> 16" do? Is there an easy way to do the mask shifting in a loop? What happens if you have multiple array positions to fill?

Do the shift before the mask because it results in smaller constants, in fact it makes them all the same as well:

Code:

---

posn[0] = (b >> 24) & 0xFF;
posn[1] = (b >> 16) & 0xFF;
...

---

If you want to put that into a loop, you need only recognise that the shift differs by exactly 8 each time.

Quote:

---

Originally Posted by 99bobster99

I can see what the mask is doing, but what does the " >> 24" and " >> 16" do? Is there an easy way to do the mask shifting in a loop? What happens if you have multiple array positions to fill?

---

The >> operator shifts the bits left that number of bits. There is a complementary << operator that shifts right that number of bits.

Yes, you can place it in a loop. You would need to make an array for the bits:

Code:

---

int shift_count[4] = { 24, 16, 8, 0 };

---

Then you can loop through. That is one way to do it. You could also start with a predetermined number and subtract 8 for each loop iteration or start at the left most byte with a shift of zero and add eight with each iteration. Many ways to accomplish this small task.

Thank you for the lesson folks, very informative! How about the shuffling of the bytes from front to back, is there an easier way to accomplish this?

Thank you Cynic and iMalc, you guys are awesome! This worked perfectly! I see the lameness in my last question, since inverting the order of the "shift_counter" array to 0, 8, 16 and 24, reversed the order of the Hex values in my Posn array.

If you do it that way around then your shift amount is just 8 time your loop counter, so you could remove the lookup table.
But hey, the small lookup table will work too.
Nice work applying some thinking to the problem.

It should be noted that your program is C++ since you are using cout. However, you seem to be reading in using scanf. Don't jumble up I/O functions from both languages, either write in C++ the C++ way or write in C the only way.

Claudiu, thank you for pointing this out! This will come in handy since I need to eventually convert this code to Linux (Fedora 13.0). I was told that as long as I keep everything in "C", it should import properly within Linux. So I am assuming, the "C++" cout command may have caused me a problem during this conversion? Is this true?

I have another challenge I am playing with, the example above converts the "Long" integer to (4) bytes within an array. How would you go the opposite way, take the (4 bytes) within the array and turn it back into a "Long" integer? I tried using the strtol command but I keep getting a "Result Too Large" error??

C and C++ are two different languages, although C is a subset of C++ in a sense. Bottom line is if you use cout, you are in the C++ beyond C realm and your code should be posted on the C++ forum.

I want to stay in the C programming realm. I will change the commands around. Thank you for pointing that out!

for interest sake, you guys showed me how to get the "Long" into (4) hex values within an array. How do you get an array of (4) hex values into a "Long"?