Copying a short through memcpy

Code:

#include <stdio.h>
#include <string.h> // memcpy
#include <iostream>

using namespace std;

int main ()
{
  char           *x = new char [20];
  unsigned short y  = 150;

  memcpy (x, (const void *)&y, 2);

  printf ("\n%d\n", x[0]);
  printf ("\n%d\n", x[1]);
}

The output of the above program is -106,

and when I allocate memory to y and then copy it using memcpy, everything works properly.

but I want to use y without allocating memory to it. What is wrong with the above program ?

Check your headers. I'm not sure about C++ (you're in the wrong forum btw), but in C, if you want to malloc() memory, you have to include stdlib.h, not string.h.

Edit: I was referring to new in your program, being approximately the same as malloc() in C. Malloc needs stdlib.h, which you did not have.

Glad you got it working, though.

Thanks for replying,

string.h has been included for memcpy !

Yes by mistake I posted this in the wrong forum, but IMO this question has some pointer logic problem and writing the same thing in C will not solve it.

The binary code of one of the 2 bytes in y is 10010110. If this byte is regarded as a signed char, its value is just -106.

BTW: A plain char might be regarded as signed char or unsigned char, depending on the compiler you are using.

Originally Posted by orientuser

The binary code of one of the 2 bytes in y is 10010110. If this byte is regarded as a signed char, its value is just -106.

BTW: A plain char might be regarded as signed char or unsigned char, depending on the compiler you are using.

Many thanks to you for the pointer !
I modified the program as follows:

Code:

#include <stdio.h>
#include <string.h> // memcpy
#include <iostream>

using namespace std;

int main ()
{
  unsigned char  *x = new unsigned char [20];
  unsigned short *y = new unsigned short [2];

  *y = 255;

  memcpy (x, (const void *)&y, 2);

  printf ("\n%d\n", x[0]);
  printf ("\n%d\n", x[1]);
}

Now the output is 48 and 32 !!
Shouldn't the output be 255 and 0 ?

memcpy (x, (const void *)&y, 2);

& is redundant.

Originally Posted by orientuser

memcpy (x, (const void *)&y, 2);

& is redundant.

I am sorry, I actually had posted the incorrect code.
The correct code is this one:

Code:

#include <stdio.h>
#include <string.h> // memcpy
#include <iostream>

using namespace std;

int main ()
{
  unsigned char  *x = new unsigned char [20];
  unsigned short y  = 255;

  memcpy (x, (const void *)&y, 2);

  printf ("\n%d\n", x[0]);
  printf ("\n%d\n", x[1]);
}

and it is working very fine.

Thanks to you

Moved to C++ programming forum.

WHY are you including iostream at all? Geez, pick a language...C or C++! If you're going to use C++ and insist on using C I/O, then use the C++ header cstdio. If you're going to use memcpy, use the C++ header cstring.

Originally Posted by rags_to_riches

If you're going to use C++ and insist on using C I/O, then use the C++ header cstdio. If you're going to use memcpy, use the C++ header cstring.

I did a man memcpy. It showed string.h. I was not aware of the cstdio and cstring headers, otherwise I would have used them instead. Thanks for pointing out !