pointer does not give memory address

[digicon]

I got through the tutorial of cplusplus.com on pointers with inserting the code:

char a;
char *b;
char **c;
a='d';
b=&a;
c=&b;

Then I wanted to print the memory addresses:

cout<<b<< endl;
cout<<c<< endl;

The terminal answered:

dP?
0xbffe4fd4

I would expect two hex numbers, but the first seems to be something different.
How can I get the memory address of char a ?
thank you
Tom

[Catfish4]

Well, when you feed a char* to std::cout, it thinks it should print a C string.

1 2 3 4 5 6 7 8 9

#include <iostream> int main() { const char *cp = "Hello, World!"; std::clog << cp << '\n'; // what should this print? std::clog << static_cast<const void *> (cp) << '\n'; }

Hello, World! 0x48f064

Traditionally... that means, back in the good old C whence C++ originated... char[] (char arrays) were used as text strings. And arrays decay to pointers. So printing a char* is a special case.

You can get around this as shown above, by using a cast.