Re: Void pointers

"Alf P. Steinbach" <>
Sun, 21 Jun 2009 15:59:13 +0200
* Marcel M?ller: wrote:

This code comes from

What a shame. It does not contain even a single character that makes use
of C++. It is C code.

/* memcpy example */
#include <stdio.h>
#include <string.h>

int main ()
  char str1[]="Sample string";
  char str2[40];
  char str3[40];
  memcpy (str2,str1,strlen(str1)+1);
  memcpy (str3,"copy successful",16);
  printf ("str1: %s\nstr2: %s\nstr3: %s\n",str1,str2,str3);
  return 0;

I'm sure it's ok but I'm concerned about the fact that memcpy takes
void* (and const void*) parameters.

Why is it unnecessary to cast str1, str2 and str3 to type char* before
the final printf statement?

? - they /are/ of type char*, so why you want to cast?

ITYM that the expressions effectively are of type 'char*' in this context.

Since that can be easily misinterpreted: the variables str1, str2 and str3 are
themselves of array type, e.g. str2 and str2 are of type 'char[40]'.

But when an array of T is used where a pointer is expected then there is an
automatic conversion to 'T*', producing a pointer to the first element.

In C that conversion always kicks in for use of an array in a run time
expression, although even in C you can do sizeof(str2) and get 40, not e.g. 4.

In C++ the conversion only kicks in where the expected type is a pointer (that
includes a simple indexing expression, and generally all C expressions that
involve arrays, since the point is to keep C compatbility), e.g. there's /no
conversion/ to 'char*' when passing str2 as argument to

    void foo( char (&t)[40 ) {}

which among other things means that if you overload foo with various array sizes
then which one's called (if any) depends on the array.


If you want an advice: throw this code away and forget about it. No C++
application should contain code like that. Dealing with raw memory is
known to be fault-prone, and C++ has dozens of ways to avoid that.
In particular do not use char*. It is likely to become the next
generation virus remote installer. Simply don't do that. Use std::string
And replace printf either by the streaming operators '<<' (ugly) or use
boost::format (similar to printf, but type safe).

Yes. :-)


- Alf

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