C++ Beginner Question

[abraxyz]

#include <iostream>
int main()
{
char buffer[3];
std::cout << (void *) buffer;
return 0;
}So, the output is something like this:
0xbfb0d89d

But, what exactly is that (void *), and how's it different from the address operator &?
Looks like a typecast to a pointer of some sort, but I still don't quite get it...
Thanks

[MadCow108]

its a type cast so the right overloaded << operator is called:
ostream & operator<<(void * str);

without the cast it will call this one:
ostream & operator<<(char * str);
as arrays convert to pointers when passed into functions

these two functions to different things, the former prints the pointer in hexadecimal format the latter prints the char's until it encounters a null terminator (and will do bad stuff when there is none)

&buffer would to the same as the cast in this case

[Jonas thomas]

#include <iostream>
int main()
{
char buffer[3];
std::cout << (void *) buffer;
return 0;
}So, the output is something like this:
0xbfb0d89d

But, what exactly is that (void *), and how's it different from the address operator &?
Looks like a typecast to a pointer of some sort, but I still don't quite get it...
Thanks
Did you intentional leave char buffer[3] uninitiated.

I'm not sure what your experimenting with the (void *)

But I tried this:

#include <iostream>
using namespace std;
int main()
{
char buffer[3];
char buffera[] ={"abc"};

cout << (void *) buffer<<endl;

cout << buffera<<endl;
cout << (void *)buffera<<endl;
return 0;
}

which produces this:


0xbf857ccd
abc
0xbf857cc9



Are you a beginner beginner, or not so beginner trying something Gucci?

[abraxyz]

Thanks for the quick replies.
Not trying anything Gucci so far : )
Was just reading a book when I encountered this, so I wasn't quite sure why the address operator wasn't applied. It was a chapter about placement new operator, by the way, illustrating dynamic memory techniques. I also tried something like (int *) buffer, and got pretty much the same result. So, I understand it has to do with the ostream << operator overloading?


&buffer would to the same as the cast in this case What would be the case where it wouldn't do the same?

EDIT:

Oh, wait, just tried with the int type:

#include <iostream>
int main()
{
int n;
std::cout << (void *) n << std::endl;
std::cout << &n;

return 0;
}

0xe77985
0xbfabdb80

So, I just want to know what is really going on here?

[abraxyz]

OK, replying to myself after the last post, lol...
n wasn't initialized, that's why I got those weird looking numbers, but after i initialized n variable, (void *) n would be the n value in hex notation. So, I finally understand the first example when I typecast the address of the first array element to a pointer. Still, doesn't seem like a particularly useful sort of typecast. But then again, I'm only a beginner...

[Jonas thomas]

OK, replying to myself after the last post, lol...
n wasn't initialized, that's why I got those weird looking numbers, but after i initialized n variable, (void *) n would be the n value in hex notation. So, I finally understand the first example when I typecast the address of the first array element to a pointer. Still, doesn't seem like a particularly useful sort of typecast. But then again, I'm only a beginner...

Think of a type cast as way of making data look like something else.
I found a bit of a discusion here.

http://www.geekinterview.com/question_details/3350
Does this make sense?
Gotta get the kid to bed.. Later

[abraxyz]

Yeah, makes a lot of sense to me now.
Thanks for the help.