Pointers/addresses

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Pointers/addresses

Pointers

I read this and I kind of understand it...

You use ted = &var to say ted points to address of var

You use *ted to say value pointed by ted

You use * var to declare a pointer of the type that precedes it.

I don't understand the arrays part...

I use the code:[php]
char array[] = "hello there";
[/php]
and it works, so why wouldn't

[php]
array = "hello there";
[/php]
not be valid? I don't understand...

Because its an array. Char only holds 1 character

Originally Posted by Auxilium

Because its an array. Char only holds 1 character

ok...

how bout
int numbers [20];
int * p;
int = p;

why would that not be valid... int holds up to 60k integers...

Originally Posted by VvITylerIvV

ok...

how bout
int numbers [20];
int * p;
int = p;

why would that not be valid... int holds up to 60k integers...

ints can hold a VALUE of around 65k.

And

[php]int = p;[/php]
is illegal. thats like saying int is p.

Originally Posted by Auxilium

ints can have a VALUE of around 65k.

And

[php]int = p;[/php]
is illegal. thats like saying Int is p.

no, int = "p"; would be saying it is p

int = p is saying int = value of p...

Originally Posted by VvITylerIvV

no, int = "p"; would be saying it is p

int = p is saying int = value of p...

int is a variable type

you cant assign a value to it

Koreans speaks the truth. The compiler won't allow that, ever...

Code:

Since pApple points to apple, when we [php]cout << *pApple;[/php] were really getting the value of apple

REMEMBER: Initialize a pointer to something or else youll have a wild pointer and can cause unwanted behavior. Even if youre not using it right away, initialize it to 0

Originally Posted by Auxilium

int is a variable type

you cant assign a value to it

Originally Posted by Void

Koreans speaks the truth. The compiler won't allow that, ever...

Well, I was just trying to ask why can't I assign a value from a variable to an array.

Originally Posted by Auxilium

Here is a simple program that creates a pointer to an integer

[php]#include <iostream>
using namespace std;

int main()
{
int apple = 5;
int *pApple = &apple;

cout << *pApple << '\n';
return 0;
}

[/php]

REMEMBER: Initialize a pointer to something or else youll have a wild pointer and can cause unwanted behavior. Even if youre not using it right away, initialize it to 0

I know how to use pointers, it was the array part that confused me... I don't understand it.

Originally Posted by Auxilium

Code:

Right, 'cause reading at address 0 definitely won't give you access violation errors.

Then why did you ask me why cant you do [php]int = p[/php]?

Originally Posted by Void

Right, 'cause reading at address 0 definitely won't give you access violation errors.

You re-assign it when youre going to use it
amiright /

Originally Posted by Auxilium

Then why did you ask me why cant you do [php]int = p[/php]?

You re-assign it when youre going to use it

because I didn't know how to explain my question then...

Originally Posted by Auxilium

Then why did you ask me why cant you do [php]int = p[/php]?

You re-assign it when youre going to use it
amiright /

Then just leave the default value wild and re-assign it later. Same results.. |:

Anyways, arrays are basically pointers..

[php]
int array[] = {1,2,3,4,5};
[/php]

To access the last integer in the array, you can either use the brackets ( arrays ).
[php]
cout << array[4];
[/php]

Or you can use pointers.

[php]
cout << *(array+4);
[/php]

Remember, the first element is always 0, so if you have 5, the last one is going to be 4. Incase you didn't already know.