Problem printing values as hex

I'm trying to print out the following 128-bit value as a sequence of 8-bit values in hexadecimal:

Code:

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unsigned int MDBuffer [4] = {0x01234567, 0x89abcdef, 0xfedcba98, 0x76543210};

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The function I'm using to print the values:

Code:

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void printOutput()
{
  unsigned char * outChars = (unsigned char*) &MDBuffer;
  for(int i = 0; i<16; i++)
  {
      printf("%02x " , *(outChars+i) );
  }
  cout<<"\n";
}

---

This produces the output:
67 45 23 01 ef cd ab 89 98 ba dc fe 10 32 54 76

I'm not sure what is happening to produce this output...

My only guess is that it relates to the 'endianess' of the values, and that my function is printing them 'backwards'.

Is this guess along the right lines, or am I making some other mistake? What could I do to correct this bug?

I'm currently researching the whole topic of endianess in C++, but it's all pretty much over my head! So any help would be greatly appreciated. :)

Quote:

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Originally Posted by Flembobs

My only guess is that it relates to the 'endianess' of the values, and that my function is printing them 'backwards'.

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Yes, that's correct.
The machine you are using is 'little endian', which means that it stores numbers in memory in lowest to highest significant byte order.

If you want to extract bytes in the correct order regardless of 'endianness' then use shifting and masking.

Code:

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unsigned int i = 0x01234567;

byte0 = (i >> 24) & 0xFF; // 0x01
byte1 = (i >> 16) & 0xFF; // 0x23
byte2 = (i >> 8) & 0xFF; // 0x45
byte3 = i & 0xFF; // 0x67

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Quote:

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Is this guess along the right lines, or am I making some other mistake? What could I do to correct this bug?

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Is there some reason you don't simply do

Code:

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for (int i = 0; i < 4; ++i)
  printf("&#37;08x ", MDBuffer[i]);

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?

My compiler wouldn't even print anything via the printf. I don't think it is a good idea to mix printf and cout in the same program. I may not have included the correct headers or for some reason the output stream wasn't being flushed in the posted program. You could just do it with with std::cout.

Code:

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int main ()
{
    unsigned int MDBuffer [4] = {0x01234567, 0x89abcdef, 0xfedcba98, 0x76543210};
    unsigned char * outChars = (unsigned char*) &MDBuffer;

    // prints backwards
    for(int i = 0; i<16; i++)
    {
        std::cout << std::hex << std::setfill('0') << std::setw(2) << static_cast<short>(*(outChars+i));
    }
    std::cout << std::endl;

  // prints forwards.  This outputs the data in the way that you wanted it.
    for(int i = 0; i<4; i++)
    {
        std::cout << std::hex << std::setfill('0') << std::setw(8) << MDBuffer[i];
    }
   
    std::cout << std::endl;

    return 0;
}

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Quote:

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Code:

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// prints backwards
for(int i = 0; i<16; i++)
{
    std::cout << std::hex << std::setfill('0') << std::setw(2) << static_cast<short>(*(outChars+i));
}

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Only for 'little endian' machines. For 'big endian' machines it will print forwards.

Thanks for your help. My eventual solution was this:

Code:

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void printOutput()
{
 
  for(int i = 0; i<4; i++)
  {
      unsigned char * outChars = (unsigned char*) &MDBuffer[i];
 
      for(int j = 3; j>-1; j--)
      {
        printf("&#37;02x " , *(outChars+j) );
      }
  }
  cout<<"\n";
}

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It's probably not optimal, and wouldn't work on a big-endian machine... but since I'm only doing this as a programming exercise, it's probably good enough!

I think I understand the problem and some possible solutions now anyways, so thanks. All this code-portability stuff drives me mad!

Quote:

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Is there some reason you don't simply do

Code:

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for (int i = 0; i < 4; ++i)
  printf("%08x ", MDBuffer[i]);

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1) I didn't think of it *facepalm*.
2) I wanted the output to be split up in groups of two to make it more readable anyways.