Pointer (adress) question.

I've been playing a bit with pointers and strings.

Code:

#include <stdio.h>

int main()
{
  char string;                        //Sets up a character named string.
  char * pString = &string;  //Makes a character pointer and assings the adress of string to it
  puts("Enter: ");
  scanf(" %s", pString);        //I know this could have been gets(), I used scanf().
  printf("%s\n",pString);
  printf("pString has been stored here: %d", pString);
  getchar();
  getchar();
  return 0;
}

Why doesn't it matter (at least as far I can see) whether I do:

Code:

  printf("pString has been stored here: %d", pString);

or

Code:

printf("pString has been stored here: %d", &pString);

The ampersand is the adress of operator in this case, so why can I get the adress without using it? Did I miss something about pointers perhaps?

pString is a pointer. The variable itself holds the address of what it points to, so of course it's going to print an address.
If you use & on pString, you will actually take the address of the pointer variable itself (not what it points to), so you'd actually get char**.
Oh and you should print pointers using %p and not %d.
Oh and gets is a very, very bad function. Never use it. And using scanf to read strings is also tricky. You need to be careful. Read my signature about it.

Hmm. I expect that crashes if you enter a string that is more than about 3 chars long.

And I would also expect that if you print pString using %d, you get one value, and if you print &pString using %d, you get a very similar value that is approximately 4 less than the pString value, because I expect the address of string to be 4 less than pString itself.

Try printing both next to each other, or subtract one from the other to see it properly.

--
Mats

Look at the declaration again:

Code:

char * pString;

This means that pString is a variable that can hold an address of a character (to wit, the first character of your string). Perhaps your string lives at address "123 Elm St." (obviously completely fictitious example). So pString = "123 Elm St.". However, pString is a variable, and must be stored somewhere in memory too, so perhaps &pString = "159 Elm St.". The value of the variable is a pointer to a string; the address of the variable is where the variable itself lives.

Originally Posted by Elysia

Oh and you should print pointers using %x and not %d.

To be perfectly correct, neither %x nor %d will be correct for pointers. If you want to use the printf format that matches exactly to a pointer, you should use %p. It will use hex, and match the size of a pointer (e.g. 64-bit on a 64-bit OS, rather than %d that is 32-bit on both 32- and 64-bit systems).

--
Mats

Wait, oops, yes, I meant %p. A typo.

Oops I overlooked that I used pString instead of string. My bad

%p for printing pointers.

So I would do:

Code:

#include <stdio.h> 
                   
int main()         
{                  
 int a = 10;       
 int * pA = &a;    
 printf("%p", pA); 
 getchar();
 return 0;        
}

This gives me a memory adress in hexadecimal system.

>printf("%p", pA);
The %p specifier expects a pointer to void. Matching the expected type is important:

Code:

printf("%p", (void*)pA);

>}
What's up with saying main will return int but not returning anything? :P

What do you call doing

Code:

(void *)pA

It looks like a sort of typecasting to me.

Originally Posted by Prelude

>printf("%p", pA);
The %p specifier expects a pointer to void. Matching the expected type is important:

Code:

printf("%p", (void*)pA);

>}
What's up with saying main will return int but not returning anything? :P

And whilst it's likely that there are systems where a void * is different from a char *, I doubt the original poster's problem will be solved by a type-cast.

And to omnificient:

It looks like a sort of typecasting to me.

Yes it is. There's nothing wrong with a typecast, although I very much doubt that it changes anything in your particular case.

--
Mats

scanf(" %s", pString); //I know this could have been gets(), I used scanf().

you have buffer of 1 char length, you really do expect some input will fit into this buffer?

and forget abour gets, when you need to read line - use fgets as described in the FAQ
http://faq.cprogramming.com/cgi-bin/...&id=1043284385

Well I'm not really familiar with C and gets() is a lot easier thatn fgets(), therefore I will use gets() and thereby ignore your advice, even if it might overwrite memory it doesn't own.

May you forever be looked on as a day one n00b, then, by your own request.

Right, what I'm saying is that I have enough to figure out already. When I find that the gets() function is no longer suitable for my needs, I'll try and figure out how the fgets() function works.

If you ignore our advice, then I shall certainly not welcome you here anymore.
You are openly ignoring advice from experts and you are choosing a function which you should never, ever use PLUS you are inviting malware vendors to hack into your application and infest others' computers.
Every programmer should be able to use fgets. It is incredibly easy, and many others have been corrected before you and use it correctly. And yes, they were all using gets before.
Now, I suggest you do the same. If not, then you should not be in the programming world or perhaps not be using C.
Never use gets!