C++: how to cast 2 bytes in an array to an unsigned short

Question

I have been working on a legacy C++ application and am definitely outside of my comfort-zone (a good thing). I was wondering if anyone out there would be so kind as to give me a few pointers (pun intended).

I need to cast 2 bytes in an unsigned char array to an unsigned short. The bytes are consecutive.

For an example of what I am trying to do:

I receive a string from a socket and place it in an unsigned char array. I can ignore the first byte and then the next 2 bytes should be converted to an unsigned char. This will be on windows only so there are no Big/Little Endian issues (that I am aware of).

Here is what I have now (not working obviously):

//packetBuffer is an unsigned char array containing the string "123456789" for testing
//I need to convert bytes 2 and 3 into the short, 2 being the most significant byte
//so I would expect to get 515 (2*256 + 3) instead all the code I have tried gives me
//either errors or 2 (only converting one byte
unsigned short myShort;
myShort = static_cast<unsigned_short>(packetBuffer[1])

Answer 1

Well, you are widening the char into a short value. What you want is to interpret two bytes as an short. static_cast cannot cast from unsigned char* to unsigned short* . You have to cast to void* , then to unsigned short* :

unsigned short *p = static_cast<unsigned short*>(static_cast<void*>(&packetBuffer[1]));

Now, you can dereference p and get the short value. But the problem with this approach is that you cast from unsigned char*, to void* and then to some different type. The Standard doesn't guarantee the address remains the same (and in addition, dereferencing that pointer would be undefined behavior). A better approach is to use bit-shifting, which will always work:

unsigned short p = (packetBuffer[1] << 8) | packetBuffer[2];

Answer 2

This is probably well below what you care about, but keep in mind that you could easily get an unaligned access doing this. x86 is forgiving and the abort that the unaligned access causes will be caught internally and will end up with a copy and return of the value so your app won't know any different (though it's significantly slower than an aligned access). If, however, this code will run on a non-x86 (you don't mention the target platform, so I'm assuming x86 desktop Windows), then doing this will cause a processor data abort and you'll have to manually copy the data to an aligned address before trying to cast it.

In short, if you're going to be doing this access a lot, you might look at making adjustments to the code so as not to have unaligned reads and you'll see a perfromance benefit.

Answer 3

The bit shift above has a bug:

unsigned short p = (packetBuffer[1] << 8) | packetBuffer[2];

if packetBuffer is in bytes (8 bits wide) then the above shift can and will turn packetBuffer into a zero, leaving you with only packetBuffer[2];

Despite that this is still preferred to pointers. To avoid the above problem, I waste a few lines of code (other than quite-literal-zero-optimization) it results in the same machine code:

unsigned short p;
p = packetBuffer[1]; p <<= 8; p |= packetBuffer[2];

Or to save some clock cycles and not shift the bits off the end:

unsigned short p;
p = (((unsigned short)packetBuffer[1])<<8) | packetBuffer[2];

You have to be careful with pointers, the optimizer will bite you, as well as memory alignments and a long list of other problems. Yes, done right it is faster, done wrong the bug can linger for a long time and strike when least desired.

Say you were lazy and wanted to do some 16 bit math on an 8 bit array. (little endian)

unsigned short *s;
unsigned char b[10];

s=(unsigned short *)&b[0];

if(b[0]&7)
{
   *s = *s+8;
   *s &= ~7;
}

do_something_With(b);

*s=*s+8;

do_something_With(b);

*s=*s+8;

do_something_With(b);

There is no guarantee that a perfectly bug free compiler will create the code you expect. The byte array b sent to the do_something_with() function may never get modified by the *s operations. Nothing in the code above says that it should. If you don't optimize your code then you may never see this problem (until someone does optimize or changes compilers or compiler versions). If you use a debugger you may never see this problem (until it is too late).

The compiler doesn't see the connection between s and b, they are two completely separate items. The optimizer may choose not to write *s back to memory because it sees that *s has a number of operations so it can keep that value in a register and only save it to memory at the end (if ever).

There are three basic ways to fix the pointer problem above:

1. Declare s as volatile.
2. Use a union.
3. Use a function or functions whenever changing types.

Answer 4

Maybe this is a very late solution but i just want to share with you. When you want to convert primitives or other types you can use union. See below:

union CharToStruct {
    char charArray[2];
    unsigned short value;
};


short toShort(char* value){
    CharToStruct cs;
    cs.charArray[0] = value[1]; // most significant bit of short is not first bit of char array
    cs.charArray[1] = value[0];
    return cs.value;
}

When you create an array with below hex values and call toShort function, you will get a short value with 3.

char array[2]; 
array[0] = 0x00;
array[1] = 0x03;
short i = toShort(array);
cout << i << endl; // or printf("%h", i);

Answer 5

You should not cast a unsigned char pointer into an unsigned short pointer (for that matter cast from a pointer of smaller data type to a larger data type). This is because it is assumed that the address will be aligned correctly. A better approach is to shift the bytes into a real unsigned short object, or memcpy to a unsigned short array.

No doubt, you can adjust the compiler settings to get around this limitation, but this is a very subtle thing that will break in the future if the code gets passed around and reused.

Answer 6

unsigned short myShort = *(unsigned short *)&packetBuffer[1];

Answer 7

static cast有不同的语法，而且你需要使用指针，你想要做的是：

unsigned short *myShort = static_cast<unsigned short*>(&packetBuffer[1]);

Answer 8

I realize this is an old thread, and I can't say that I tried every suggestion made here. I'm just making my self comfortable with mfc, and I was looking for a way to convert a uint to two bytes, and back again at the other end of a socket.

There are alot of bit shifting examples you can find on the net, but none of them seemed to actually work. Alot of the examples seem overly complicated; I mean we're just talking about grabbing 2 bytes out of a uint, sending them over the wire, and plugging them back into a uint at the other end, right?

This is the solution I finally came up with:

class ByteConverter
{
public:
 static void uIntToBytes(unsigned int theUint, char* bytes)
  {
   unsigned int tInt = theUint;

   void *uintConverter = &tInt;
   char *theBytes = (char*)uintConverter;

   bytes[0] = theBytes[0];
   bytes[1] = theBytes[1];
  }
 static unsigned int bytesToUint(char *bytes)
  {
   unsigned theUint = 0;

   void *uintConverter = &theUint;
   char *thebytes = (char*)uintConverter;

   thebytes[0] = bytes[0];
   thebytes[1] = bytes[1];

   return theUint;
  }
};

Used like this:

unsigned int theUint;
char bytes[2];
CString msg;

ByteConverter::uIntToBytes(65000,bytes);
theUint = ByteConverter::bytesToUint(bytes);

msg.Format(_T("theUint = %d"), theUint);
AfxMessageBox(msg, MB_ICONINFORMATION | MB_OK);

Hope this helps someone out.

Answer 9

Did nobody see the input was a string!

/* If it is a string as explicitly stated in the question.
 */
int byte1 = packetBuffer[1] - '0'; // convert 1st byte from char to number.
int byte2 = packetBuffer[2] - '0';

unsigned short result = (byte1 * 256) + byte2;

/* Alternatively if is an array of bytes.
 */
int byte1 = packetBuffer[1];
int byte2 = packetBuffer[2];

unsigned short result = (byte1 * 256) + byte2;

This also avoids the problems with alignment that most of the other solutions may have on certain platforms. Note A short is at least two bytes. Most systems will give you a memory error if you try and de-reference a short pointer that is not 2 byte aligned (or whatever the sizeof(short) on your system is)!

Answer 10

char packetBuffer[] = {1, 2, 3};
unsigned short myShort = * reinterpret_cast<unsigned short*>(&packetBuffer[1]);

I (had to) do this all the time. big endian is an obvious problem. What really will get you is incorrect data when the machine dislike misaligned reads! (and write).

you may want to write a test cast and an assert to see if it reads properly. So when ran on a big endian machine or more importantly a machine that dislikes misaligned reads an assert error will occur instead of a weird hard to trace 'bug' ;)

Answer 11

在Windows上，您可以使用：

unsigned short i = MAKEWORD(lowbyte,hibyte);