Thread: Newbie question (Pointers)

It is indeed a pointer to an array of char of 256 elements. Each increment of that pointer would skip to the next array.
Use a pointer to char to point to the first element in the array. When incrementing it, it will step forward to the next element, because that's how pointers work. They step forward 1 element.

Why do you declare array b?
I think i understand that PArray is a Pointer to an array of 256 components. I mean, in the declaration the compiler allocate 256 bytes of memory. I don't see the need of declaring a new array of 256 components...

No, it does not allocate anything. It is a pointer. See the article I referenced.

Ok, then why if i do

Code:

#include <stdio.h>
#include <stdlib.h>

char (*PArray)[256];
char *p;

int main()
{
    p = &PArray;
    *p = 'H';
    *p++;
    *p = 'i';
    *p++;
    *p = '!';
    *p = '\0';
    p = &PArray;
    printf("&#37;s \n", p);
    return 0;
}

Works fine. Sorry if it's a stupid question but i really want to understand what is happening.

I believe that you are just lucky (or unlucky, as the case may be): as far as I can tell, you are dereferencing a null pointer when you write *p since p is statically allocated, hence it starts off as a null pointer.

Actually, why are PArray and p global variables?

You would like that, don't you?
That code is horribly broken it won't even compile on a C++ compiler.
It is undefined behavior x1000.

And for the record, it crashes for me.

Time for pointer materials (again):
http://cslibrary.stanford.edu/104/
http://cboard.cprogramming.com/showp...3&postcount=31
http://cboard.cprogramming.com/showp...65&postcount=4
http://cpwiki.sourceforge.net/A_pointer_on_pointers

p = &PArray;
Assigns the address of the actual pointer to p. Likely not what you want.
And, it... should... not... WORK!

*p = 'H';
Stores 'H' in the first byte of the pointer PArray.

*p++;
Dereferences p (no effect), then increments the pointer p by one byte, to point to PArray + 1.

*p = 'i';
Assigns 'i' to PArray + 1.

*p++;
Same as above. Now points to PArray + 2.

*p = '!';
Assigns '!' to PArray + 2.

*p = '\0';
Assigns '\0' to PArray + 2.

p = &PArray;
Same as before.

Do you pay heed to warnings at all? You should get a multitude of warnings.
Try printing the value of PArray at the end of the program, and at the beginning:

Code:

printf("%p", PArray);

Originally Posted by Tux_Fan

You were right, that's horrible. Now i understand a little more. I don't have any warnings, and all compiler warnings are enabled!
I'm replacing the value of the pointer, and the consecutive bytes, with chars! What a mess!

Now, i don't understand why i have to declare that PArray is a pointer to an array of 256 components.
Why is the difference if i declare PArray like a pointer to an array of char?

The problems with your code was as follows:

In the first one, your created a pointer to an array of chars, with 256 elements, but again, you only created the pointer, so it pointed to nothing (did you see the pointer tutorial movie?).
So in my example, I created a local array and set the pointer to point to it. So it works.

In your second example, it worked because you used the PArray pointer as a buffer. If you would have done p = &*PArray[0], and then used the code as usual, it would crash, and then it would be equal to the same code as before.

It all boils down to that you must have a storage first. A pointer is not a storage - it is merely a variable pointing to something, whether that is valid or not.

Originally Posted by Tux_Fan

I don't have any warnings, and all compiler warnings are enabled!

That is weird. What version of gcc are you using?

Originally Posted by Tux_Fan

Now, i don't understand why i have to declare that PArray is a pointer to an array of 256 components.

You don't. It depends on what you want to do.

Originally Posted by Tux_Fan

Why is the difference if i declare PArray like a pointer to an array of char?

What do you mean?

Originally Posted by Elysia

http://cpwiki.sourceforge.net/Common...llocating_them

Code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
	char b[256];
	char (*PArray)[256] = &b;
	char* p = &*PArray[0];
	*p = 'a';
	printf("%c \n", *p);
	return 0;
}

Works as expected.

Is it okay to simply write that as,
char* p = PArray[0];