C++ pointer Question

jackdaw

Hi folks ,

I have a question about pointers.
From this tutorial
http://www.cplusplus.com/doc/tutorial/pointers.html

it has

char * terry = "hello"; 

It is important to indicate that terry contains the value 1702, and not 'h' nor "hello", although 1702 indeed is the address of both of these.

************************
The use of 1702 was just an example for an address.
but my problem is when i do this and then cout terry this code exactly

char * terry = "hello";

cout<<terry<<endl;

it outputs "hello" !!!
not the address , *terry outputs "h" which makes sense ...

but to get the address i need to cout &terry

marco_polo

jackdaw wrote:

Hi folks ,

I have a question about pointers.
From this tutorial
http://www.cplusplus.com/doc/tutorial/pointers.html

it has

char * terry = "hello"; 

It is important to indicate that terry contains the value 1702, and not 'h' nor "hello", although 1702 indeed is the address of both of these.

************************
The use of 1702 was just an example for an address.
but my problem is when i do this and then cout terry this code exactly

char * terry = "hello";

cout<<terry<<endl;

it outputs "hello" !!!
not the address , *terry outputs "h" which makes sense ...

but to get the address i need to cout &terry

More of a java man its been a while since I done C++ but I think I know.

terry does contain the memory address value such as 1702. However in standard C\C++ char* arrays are treated diferently by stdout functions suct as cout. It assumes a that it is a null terminated string and prints all the chars untill it hits a null.

If it was of type int* for example the actual memory address would be printed

jackdaw

yes i was thinking it had to do with it being Char, with ints it prints the address

Cake Fiend

If you used something like

printf("The value of 'terry' is: %d\n", terry);

You should get the memory address.

Creamy Goodness

would it not be %p - for pointer - instead of %d as the address will more than likely be in hex

pid()

It would actually be %s for character. The difference is in printing the pointers memory address, or the contents of the pointer (& or omitting it).

A pointer variable is declared by giving it a type and a name (e.g. int *ptr) where the asterisk tells the compiler that the variable named ptr is a pointer variable and the type tells the compiler what type the pointer is to point to (integer in this case).
Once a variable is declared, we can get its address by preceding its name with the unary & operator, as in &ptr.

For instance, if j is a char:

printf("j has the value %s and is located at memory address %s", j, &j);

You can also use %p like so:

    printf("%p\n", (void *)j);

%p is for pointers of type (void *), so if j isn't a void * pointer then you need to typecast it, as shown above.

%x prints it as hex (0xce01d4)
%d prints the memory address as decimal (123)
%p prints it as hex (0xce01d4)
%s also prints it as hex.

It really depends on the format you want to print the address as. You could even use %u.