My Pointer Test Result

HelloWorld · #1 (**permalink**) 09-19-2007, 01:32 AM

Here's the code that I use to test pointer:

PHP Code:

  #include <iostream>

using namespace std;

int main() {
    int num = 5;
    cout << num << endl;
    cout << &num << endl;

    int *tester = &num;

    cout << *tester << endl;

    *tester = 6;

    cout << *tester << endl;
    cout << num << endl;

    int *tester2 = tester;

    cout << tester2 << endl;
    cout << *tester2 << endl;
}

So can I conclude that pointer CAN'T be a variable, it's only something that can be used to reference the other variable's memory. And when we change this pointer's value to something else, the variable that is pointed to is also changed because this is passed by reference...? Feel free to add some more summary that I can use to get better understanding of how does the pointer works. Moreover, I need to use * sign whenever I use pointer throughout my program..? If I don't use it, it's just going to be the memory of that pointer, so if I want to make another pointer to this pointer, I should do the last step.. (I DON'T NEED TO USE & sign) lol.. Let me know if I missed any test in this process.. thanx

Lee · #2 (**permalink**) 09-19-2007, 02:35 AM

For more Explanation check out... Pointers That has all the information and explanation you will need on pointers and displays easy pictures so you can see how they work.

I tend to use pointers when i need a user defined array, that being array[i] and i is the value from the user, obviously that can not be done in C++ so i use pointers too do that instead.

HelloWorld · #3 (**permalink**) 09-20-2007, 01:01 PM

Above image is pretty nice clarification

Lee · #4 (**permalink**) 09-20-2007, 02:33 PM

Quote:

Originally Posted by HelloWorld

Above image is pretty nice clarification

Those are the images i was on about, they make it really clear so when you make your own examples its easy!

Bench · #5 (**permalink**) 09-27-2007, 03:16 AM

Quote:

Originally Posted by HelloWorld

So can I conclude that pointer CAN'T be a variable, it's only something that can be used to reference the other variable's memory. And when we change this pointer's value to something else, the variable that is pointed to is also changed because this is passed by reference...? Feel free to add some more summary that I can use to get better understanding of how does the pointer works. Moreover, I need to use * sign whenever I use pointer throughout my program..? If I don't use it, it's just going to be the memory of that pointer, so if I want to make another pointer to this pointer, I should do the last step.. (I DON'T NEED TO USE & sign) lol.. Let me know if I missed any test in this process.. thanx

You are correct in your assertion that a pointer is used to reference other variables or objects in a program, and that modifying a pointer causes a different variable or object to be referenced.

As for your statement "A pointer can't be a variable" - I assume you mean that The identifier (name) for the pointer cannot be used to directly manipulate the referenced variable. Which, again, is correct. The de-reference operator ( * ) must be used to access the referenced variable.

However, in general, a pointer occupies space in memory, and has a data type, therefore a pointer certainly can be a variable itself - It is possible to create a pointer-to-pointer

Code:

int i            = 5;        //type: int
int* p_i         = &i;       //type: pointer (-to-int)
int** p_p_i      = &p_i;     //type: pointer (-to-pointer(-to-int) )

Its important to get out of the mindset where you think of a pointer as being some kind of a mutator for a variable (Which is how alot of people seem to start out), and begin to treat a pointer as a whole new type.

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HelloWorld (09-23-2007)