Thread: pointers *shudder*

I believe i have a fair understanding of pointers. What i don't understand or rather if i'm under estimating the powerfulness of pointers. So basically we can by-pass call by value restrictions, they can be used as part of linked lists but what else?

I have come to conclusion that pointers offer the following benefits?

- We can dynamically modify whatever the pointer is pointing to. So for example we could have a method which does something and modify the contents of the pointed value, we can then print the value in the main method by de-referencing the pointer and finding out what its pointing *TO* (i.e. *xPtr = 2) makes x =2 if *xPtr = &x (if *xPtr points to X), rather than using global variables (which are bad).

- They're the equivalent of getters and setters in java (or kind of??)

And i've also written something small to demonstrate my basic understanding. Please correct me if i'm wrong:

Code:

#include <stdio.h>
void calculate(int *xPtr);
int main()
{
	int x = 0;
	int *xPtr; // Create a pointer and a memory location is allocated.
	
	xPtr = &x;
	*xPtr = 2;
	//calculate(&x); // Pass a reference to the variable x, which will make xPtr point to x
	
	printf("%d", x); // print what x contains
	printf("%d", *xPtr); // print what xPtr is pointing to in this case x.
	
	return 0;
}

void calculate(int *xPtr)
{
	*xPtr = 2; //set whatever xPtr is pointing to (x) to 2.
}

Also what stumps me is the following:

If i pass a reference of x to calculate it makes xptr point to X and sets xptr to 2 which makes X 2 since xptr is pointing to x now. But for some reason i can't de-reference *xPtr and print it the program just crashes ?

Code:

#include <stdio.h>
void calculate(int *xPtr);
int main()
{
	int x = 0;
	int *xPtr; // Create a pointer and a memory location is allocated.
	
	//xPtr = &x;
	calculate(&x); // Pass a reference to the variable x, which will make xPtr point to x
	
	printf("%d", x); // print what x contains
	printf("%d", *xPtr); // print what xPtr is pointing to in this case x.
	
	return 0;
}

void calculate(int *xPtr)
{
	*xPtr = 2; //set whatever xPtr is pointing to (x) to 2.
}

You forgot about dynamic memory allocation, which is probably the most important function of the pointer. This is why your second example is broken.

>But for some reason i can't de-reference *xPtr and print it the program just crashes ?
You're trying to dereference an uninitialized pointer. Before you can dereference a pointer, it must point to memory that you own, either by assigning the address of an existing object or allocating dynamic memory.

>sorry i meant xPtr = &x;
You failed to do that in the code that had the problem. To be more specific, you commented it out:

Code:

int *xPtr;
	
//xPtr = &x;
calculate(&x); /* xPtr is still uninitialized here */

Originally Posted by Axel

what confuses me is that:

void calculate(int *xPtr)


i'm passing an &x to it. When the arguments accept a pointer (i.e. xPtr is pointing to x) Why is that?

Huh? I don't understand what your question is.

Originally Posted by Axel

what confuses me is that:

void calculate(int *xPtr)


i'm passing an &x to it. When the arguments accept a pointer (i.e. xPtr is pointing to x) Why is that?

xPtr is a pointer which accepts the address of an integer. x is an integer, so you pass the address of x (&x) to the pointer.

&x generates a pointer, a pointer to x.

Actually it generates the address of a given variable.


Quzah.