Thread: pointers to multidimensional arrays

Who is to say that memory layout, pointers & all that isn't confusing?
Hey, everything is confusing in a way.

Not I. But that's not to say that it's something that shouldn't be learned early. I skipped most of that (pointers, memory stuff), in class when I first started programming and to this day I sometimes have to think about how to use and implement them properly in code for a little bit. I shouldn't have to, but there you go. So from my experience I figure the earlier you know it the better.

Edit: As you said. There's a difference in the school of thought. I don't think you're going to convince me one way's better than the other here. I never did consider learning about containers before arrays and other such nuances, but that's because it was how I was taught.

From my experience, before I learned all the advanced stuff and techniques available with C++, I did the old "C" way, with overly complex code that even sometimes relied on certain implementation.
Had I learned of these "higher" level things first, I probably would never have done that in the first place and the old code would have been better.

And again I am furious with C++. There is no way of predicting how to point to something. I would never in my lifetime guess

Code:

int (*)[3][2] p = x;

that (*)... is that a typecast?

I know the index of first dimension is optional but the parentheses with deferencing operator? And why do dimensions go before the pointer name
when in an array declaration they go after?

I have issues with pointers! function pointers, array pointers, all pointers.

Was Bjarne Strostrup smoking crack when he was developing the syntax?

I swear to almighty one i'll start learning calculus instead of C++

>> And why do dimensions go before the pointer name when in an array declaration they go after?
Because it's a pointer to the array. The array's memory is already there. You have to specify that the pointer points to that memory. You know that int *ptr; makes makes a pointer called ptr. The syntax extends beyond that in case you want to point to an array. There's no sense putting them after the variable name. That'll cause an array of pointers. Make sense? It's just something you have to wrap your head around. This is useful guide about stuff (relevant part copied below).

5. Using the variable a, give definitions for the following:
a) An integer
b) A pointer to an integer
c) A pointer to a pointer to an integer
d) An array of 10 integers
e) An array of 10 pointers to integers
f) A pointer to an array of 10 integers
g) A pointer to a function that takes an integer as an argument and returns an integer
h) An array of ten pointers to functions that take an integer argument and return an integer

The answers are:
a) int a; // An integer
b) int *a; // A pointer to an integer
c) int **a; // A pointer to a pointer to an integer
d) int a[10]; // An array of 10 integers
e) int *a[10]; // An array of 10 pointers to integers
f) int (*a)[10]; // A pointer to an array of 10 integers
g) int (*a)(int); // A pointer to a function a that takes an integer argument and returns an integer
h) int (*a[10])(int); // An array of 10 pointers to functions that take an integer argument and return an integer

In short, it's like how int **ptrptr; is a pointer to a pointer, but the second star is replaced by an array specifier.

>> I have issues with pointers! function pointers, array pointers, all pointers.
All about pointers. I can't remember if it does function pointers or not...

>> I swear to almighty one i'll start learning calculus instead of C++
Do both and you can be cool too!

Edit: In fact, I believe Elysia's syntax is incorrect... it should be int (*p)[3][2], I believe, and not int (*)[3][2] p = x;e.g.:

Code:

int main( void ) {
  int myarray[4][3][2] = { NULL };
  int (*p)[3][2] = &myarray[0];

  std::cout<< myarray[0][0][0] << "\n";
  *p[0][0] = 42;
  std::cout<< myarray[0][0][0];
   
  return 0;
}

Originally Posted by Luciferek

And again I am furious with C++. There is no way of predicting how to point to something. I would never in my lifetime guess

Code:

int (*)[3][2] p = x;

that (*)... is that a typecast?

It's the pointer marker! It says it's a pointer.

I have issues with pointers! function pointers, array pointers, all pointers.

You don't necessarily need to use them that often in C++, though :)

Was Bjarne Strostrup smoking crack when he was developing the syntax?

You're blaming the wrong individual. It's a relic from C, so blame C!

Originally Posted by twomers

Edit: In fact, I believe Elysia's syntax is incorrect... it should be int (*p)[3][2], I believe, and not int (*)[3][2] p = x;e.g.:

Code:

int main( void ) {
  int myarray[4][3][2] = { NULL };
  int (*p)[3][2] = &myarray[0];

  std::cout<< myarray[0][0][0] << "\n";
  *p[0][0] = 42;
  std::cout<< myarray[0][0][0];
   
  return 0;
}

Of yes, you're right.
Dang. Didn't check that. It's not very often I use pointers to arrays.

Originally Posted by Elysia

From my experience, before I learned all the advanced stuff and techniques available with C++, I did the old "C" way, with overly complex code that even sometimes relied on certain implementation.
Had I learned of these "higher" level things first, I probably would never have done that in the first place and the old code would have been better.

Oh yes, I just saw it now.
Well, to be honest, I don't know which way is "right," so I don't really have any convincing to do! :)

OK..
I can live with

Code:

int (*p)[3][2] = x;

It is somewhat consistent with other pointer declarations.

But than again... how do i figure out when do I need parentheses when declaring pointers in general?
When declaring a pointer to one dimensional array I don't need (* )
I also noticed similar thing with function pointers where you put (* ) around the
name of the pointer!

You need parentheses, you just apparently haven't been using them.

well no... I do not need parentheses for one dimentional arrays

Code:

int main(void){
int v[3]={3,2,3};
int *b=v;
cout<< b[1];
cin.get();
}

But..

Code:

int main(void){
int v[3][2]={3,2,3,3,2,3};
int *b [2] =v;
cout<< b[1][2];
cin.get();
}

Gives an error

Originally Posted by tabstop

...Normally the * will bind with the type (int in this case), rather than the name.

But your code seems to suggest otherwise!
Another good reason to use T* var instead of T *var.

So similarly

Code:

int *bob(int x); //a function taking int returning pointer-to-int
int (*bob2)(int x); // a pointer to a function taking int and returning int
int *(*bob3)(int x); // a pointer to a function taking int and returning pointer-to-int

Code:

int* bob(int x); //a function taking int returning pointer-to-int
int (* bob2)(int x); // a pointer to a function taking int and returning int
int* (* bob3)(int x); // a pointer to a function taking int and returning pointer-to-int

See? See!?! Much easier to read!

Originally Posted by Luciferek

well no... I do not need parentheses for one dimentional arrays

Code:

int main(void){
int v[3]={3,2,3};
int *b=v;
cout<< b[1];
cin.get();
}

This isn't a pointer to an array.

Code:

int main()
{
    int v[3]={3,2,3};
    int (* b)[3] = v;
    cout<< b[1];
    cin.get();
}

But this is!
And please - drop the void from main. It's not necessary in C++.

Code:

int main()
{
    int v[3]={3,2,3};
    int (* b)[3] = v;
    cout<< b[1];
    cin.get();
}

gives me an error

error: cannot convert `int*' to `int (*)[3]' in initialization

Ah yes, you have to do &v instead of v.
Dang, I need to test my code more before posting it.

Code:

int main()
{
    int v[3]={3,2,3};
    int (* b)[3] = &v;
    cout<< b[1];
    cin.get();
}

There. Now it compiles.

Man this is confusing...

Eh, if b is a pointer to a lone array, then printing b[1] results in undefined behaviour, since you are accessing the next array, but it does not exist.