pointers *shudder*

This is a discussion on pointers *shudder* within the C Programming forums, part of the General Programming Boards category; I believe i have a fair understanding of pointers. What i don't understand or rather if i'm under estimating the ...

  1. #1
    Learner Axel's Avatar
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    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.
    }
    Last edited by Axel; 08-28-2006 at 06:55 AM.

  2. #2
    Devil's Advocate SlyMaelstrom's Avatar
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    You forgot about dynamic memory allocation, which is probably the most important function of the pointer. This is why your second example is broken.
    Last edited by SlyMaelstrom; 08-28-2006 at 06:59 AM.
    Sent from my iPadŽ

  3. #3
    Code Goddess Prelude's Avatar
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    >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.
    My best code is written with the delete key.

  4. #4
    Registered User Rennor's Avatar
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    I am thinking about your compilers warning level, it should have warned you about using uninitialized variable.

  5. #5
    Learner Axel's Avatar
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    Quote Originally Posted by Prelude
    either by assigning the address of an existing object or allocating dynamic memory.
    x is an exisiting object i'm also assigning it &x and pass it to calculate which accepts a pointer.

    Isn't it the same as :

    *xPtr = &x;

  6. #6
    Learner Axel's Avatar
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    so:

    Code:
    #include <stdio.h>
    void calculate(int *xPtr);
    int main()
    {
    	int x = 0;
    	int *xPtr = &x; // Make xPtr point to x.
    	
    	calculate(&x); // Pass a reference to the variable x, which will make xPtr point to x??? OR
    	
    	printf("%d", x); // print what x contains
    	printf("%d", *xPtr); // print what x contains
    	
    	return 0;
    }
    
    void calculate(int *xPtr)
    {
    	*xPtr = 2; //set whatever xPtr is pointing to (x) to 2.
    }
    it still defeats the purpose of calculate accepting a memory reference...

  7. #7
    Registered User Rennor's Avatar
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    Quote Originally Posted by Axel
    Isn't it the same as :

    *xPtr = &x;
    That would be all wrong since *xPtr while being uninitialized is pointing "nowhere" or "somewhere" And you try to assign it value which is reference of x.

  8. #8
    Learner Axel's Avatar
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    Quote Originally Posted by Rennor
    That would be all wrong since *xPtr while being uninitialized is pointing "nowhere" or "somewhere" And you try to assign it value which is reference of x.

    sorry i meant xPtr = &x;

  9. #9
    Learner Axel's Avatar
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    ok i'm guessing:

    Code:
    void calculate(int *xPtr)
    {
    	*xPtr = 2; //set whatever xPtr is pointing to (x) to 2.
    }
    this only sets whatever xPtr is pointing to before this it's not pointing to no (i.e unalloacted reference) where so therefore is the reason why it crashed.

  10. #10
    Code Goddess Prelude's Avatar
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    >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 */
    My best code is written with the delete key.

  11. #11
    Learner Axel's Avatar
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    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?

  12. #12
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    Quote 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.
    If you understand what you're doing, you're not learning anything.

  13. #13
    Devil's Advocate SlyMaelstrom's Avatar
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    Quote 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.
    Sent from my iPadŽ

  14. #14
    Frequently Quite Prolix dwks's Avatar
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    &x generates a pointer, a pointer to x.
    dwk

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  15. #15
    ATH0 quzah's Avatar
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    Actually it generates the address of a given variable.


    Quzah.
    Hope is the first step on the road to disappointment.

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