Printing address of a char with a pointer

[joemolot]

Hey, why does this code return a set of really weird squigly characters as the address of a variable of type char?

#include <iostream>
#include <string>
#include <conio.h>

int main(){

	std::cout << "Pointers example:" << std::endl;
	_getch();
	std::cout << "Enter a character to store as a variable. " << std::endl;
	std::cout << "If a string is entered the first letter only will be stored: " << std::endl;
	char var;
	std::cin >> var;
	std::cout << "You entered '" << var << "'." << std::endl;
	char* varloc;
	varloc = &var;
	std::cout << "It is being stored at: " << varloc << "." << std::endl;
	std::cout << std::endl;
	std::cout << std::endl;
	
}

[Yanson]

Line 16 you need a * before varloc to dereferance it such as *varloc

[thejman250]

- No he doesn't. Doing so will display the value at the address of "varloc", which is var.

- What you have should work correctly AFAIK.

[joemolot]

now it just prints the value of the variable?

I rewrote line 16 like this:

std::cout << "It is being stored at: " << *varloc << "." << std::endl;

I enter 'f' and it says:

'It is being stored at: f.'

[joemolot]

http://imgur.com/ilSBHI6

Here's a picture

[thejman250]

- That's because you had it correct the first time. That user that said to dereference the variable was incorrect.

[thejman250]

- Why it's displaying special characters, i'm not sure. However, the method you are using is correct AFAIK.

[joemolot]

ok, thanks, i've gone back to my original code.

i also just wrote

char letter = 's';
cout << &s << endl;

and got the same weird characters, perhaps it's something to do with visual studio?

[thejman250]

- I'm not sure, as it just happened in codeblocks for me.

- I'll probably make a project and try it again.

[Zhuge]

No, cout is simply overloaded to presume char*'s are C-style strings and print the contents rather than the address. To get the address stored in the char*, cast it to another pointer type then print it out.

[Chervil]

#include <iostream>

int main()
{
	std::cout << "Pointers example:" << std::endl;

	char var = 'A';
	char* varloc = &var;
	
	std::cout << "  (char *)  : " <<  (char *) varloc <<  std::endl;
	std::cout << "  (void *)  : " <<  (void *) varloc <<  std::endl;

    return 0;	
}

Output:

Pointers example:
  (char *)  : A; #
  (void *)  : 0x23ff3b

[joemolot]

Chervil would you be able to explain that code to me?
cheers

[Chervil]

It's just an example of what Zhuge explained above, "cast it to another pointer type then print it out."

[joemolot]

i meant how you wrote
(char *) and it didn't work, but then you wrote (void *) and it did?
why is that?

[Chervil]

The code i wrote was a bit odd. I put (char *) simply for comparison purposes. In fact if that part is omitted, the behaviour is exactly the same, since the variable was already a pointer to char.

It didn't work because, as Zhuge said, cout is overloaded to presume char*'s are C-style strings and print the contents rather than the address.

The word "overloaded" here is important. It refers to a function which has multiple versions with the same name but a different type of parameter.

For example, you could have two functions called add:

int add(int x, int y) { return x + y ; }
double add(double x, double y) { return x + y ; }

When you call the function, the compiler chooses which one to call, depending upon the type of parameter.

In the case of cout << if the parameter is a pointer to char, the compiler assumes this is a character string, and so tries to print out the string.

If the parameter is a pointer to void (that is, a general pointer to no specific type), the compiler chooses a different version, which prints the value of the pointer itself.

[joemolot]

so it takes the address of the variable as a string, as oppose to just printing out the address?

thanks for the help :)