help with dynamic array structure.

[Steven Taylor]

I'm learning C++ from C++ Primer Plus 5th edition. I'm struggling at
the moment with declaring an array of three structures by using 'new' to
allocate memory. (this is one of the programming exercises at the end
of each chapter). My attempt is as per below. I get a compile error
indicating that -

"base operand of '->' has no pointer type 'candybar' "

// Q-09.cpp -- Question 9
#include <iostream>
struct candybar
{
//char brand[20];
std::string brand; //in this context - must use std::string
float weight;
int calories;
};
int main()
{
using namespace std;
candybar * snack = new candybar[3];

snack[0]->brand = "Mocha Munch"; // above error starts here
snack[0]->weight = 2.3;
snack[0]->calories = 350;

snack[1]->brand = "Crunchie";
snack[1]->weight = 1.9;
snack[1]->calories = 400;

snack[2]->brand = "Mars Bar";
snack[2]->weight = 4.5;
snack[2]->calories = 200;

cout << endl;
cout << "Snack brand : " << snack[0]->brand << endl;
cout << "Snack weight : " << snack[0]->weight << endl;
cout << "Snack calories : " << snack[0]->calories << endl;

cout << "Snack brand : " << snack[1]->brand << endl;
cout << "Snack weight : " << snack[1]->weight << endl;
cout << "Snack calories : " << snack[1]->calories << endl;

cout << "Snack brand : " << snack[2]->brand << endl;
cout << "Snack weight : " << snack[2]->weight << endl;
cout << "Snack calories : " << snack[2]->calories << endl;

delete snack;
return 0;
}

Any help would be appreciated, I really can't see why this shouldn't
work. I read the chapter umpteen times but I must be missing something.

Steve Taylor
Melbourne, Australia.

[John Carson]

"Steven Taylor" <ne******@super jacent.net> wrote in message
news:42******** *************** @news.optusnet. com.au

I'm learning C++ from C++ Primer Plus 5th edition. I'm struggling at
the moment with declaring an array of three structures by using 'new'
to allocate memory. (this is one of the programming exercises at
the end of each chapter). My attempt is as per below. I get a
compile error indicating that -

"base operand of '->' has no pointer type 'candybar' "

// Q-09.cpp -- Question 9
#include <iostream>
struct candybar
{
//char brand[20];
std::string brand; //in this context - must use std::string
float weight;
int calories;
};
int main()
{
using namespace std;
candybar * snack = new candybar[3];

snack[0]->brand = "Mocha Munch"; // above error starts here
snack[0]->weight = 2.3;
snack[0]->calories = 350;

Replace all of the -> operators with the . operator.

snack is a pointer, but using the subscript operator dereferences the
pointer. Thus snack[0] is of type candybar, not pointer to candybar.
--
John Carson

[Steven Taylor]

John Carson wrote:

"Steven Taylor" <ne******@super jacent.net> wrote in message
news:42******** *************** @news.optusnet. com.au

I'm learning C++ from C++ Primer Plus 5th edition. I'm struggling at
the moment with declaring an array of three structures by using 'new'
to allocate memory. (this is one of the programming exercises at
the end of each chapter). My attempt is as per below. I get a
compile error indicating that -

"base operand of '->' has no pointer type 'candybar' "

// Q-09.cpp -- Question 9
#include <iostream>
struct candybar
{
//char brand[20];
std::string brand; //in this context - must use std::string
float weight;
int calories;
};
int main()
{
using namespace std;
candybar * snack = new candybar[3];

snack[0]->brand = "Mocha Munch"; // above error starts here
snack[0]->weight = 2.3;
snack[0]->calories = 350;

Replace all of the -> operators with the . operator.

snack is a pointer, but using the subscript operator dereferences the
pointer. Thus snack[0] is of type candybar, not pointer to candybar.

Thanks for your hlep John, it certainly compiled ok.

Steve.

[Karl Heinz Buchegger]

Steven Taylor wrote:

I'm learning C++ from C++ Primer Plus 5th edition. I'm struggling at
the moment with declaring an array of three structures by using 'new' to
allocate memory. (this is one of the programming exercises at the end
of each chapter). My attempt is as per below. I get a compile error
indicating that -

"base operand of '->' has no pointer type 'candybar' "

// Q-09.cpp -- Question 9
#include <iostream>
struct candybar
{
//char brand[20];
std::string brand; //in this context - must use std::string
float weight;
int calories;
};
int main()
{
using namespace std;
candybar * snack = new candybar[3];

snack[0]->brand = "Mocha Munch"; // above error starts here

Your problem is here.
It is the way array indexing is defined in C++

a[index] is equivalent to *(a_as_ptr+inde x)

where a_as_ptr denotes a pointer to the first element of the array.

What you wanted to write was:

snack pointer
snack+0 index offset of index 0 to that pointer
*(snack+0) object at the index offset to that pointer
(*(snack+0)).br and the brand component of the object at the index
offset to that pointer

Due to the way array indexing is defined you can replace parts of the last expression.

(*(snack+0)).br and is equivalent to

snack[0].brand

So in short:
Even if snack is a pointer, snack[0] is no longer of pointer type, but is already
the object in the array. Therefore you don't use ->, but simply access the component
directly with a '.' operation.

--
Karl Heinz Buchegger
kb******@gascad .at

[Steven Taylor]

Karl Heinz Buchegger wrote:
snip

Your problem is here.
It is the way array indexing is defined in C++

a[index] is equivalent to *(a_as_ptr+inde x)

where a_as_ptr denotes a pointer to the first element of the array.

What you wanted to write was:

snack pointer
snack+0 index offset of index 0 to that pointer
*(snack+0) object at the index offset to that pointer
(*(snack+0)).br and the brand component of the object at the index
offset to that pointer

Due to the way array indexing is defined you can replace parts of the last expression.

(*(snack+0)).br and is equivalent to

snack[0].brand

So in short:
Even if snack is a pointer, snack[0] is no longer of pointer type, but is already
the object in the array. Therefore you don't use ->, but simply access the component
directly with a '.' operation.

Thanks, much appreciated.

Steve Taylor

[yinquan]

"John Carson" <jc************ ****@netspace.n et.au> ????
news:42******** *************** @un-2park-reader-02.sydney.pipen etworks.com.au. ..

"Steven Taylor" <ne******@super jacent.net> wrote in message
news:42******** *************** @news.optusnet. com.au

I'm learning C++ from C++ Primer Plus 5th edition. I'm struggling at
the moment with declaring an array of three structures by using 'new'
to allocate memory. (this is one of the programming exercises at
the end of each chapter). My attempt is as per below. I get a
compile error indicating that -

"base operand of '->' has no pointer type 'candybar' "

// Q-09.cpp -- Question 9
#include <iostream>
struct candybar
{
//char brand[20];
std::string brand; //in this context - must use std::string
float weight;
int calories;
};
int main()
{
using namespace std;
candybar * snack = new candybar[3];

snack[0]->brand = "Mocha Munch"; // above error starts here
snack[0]->weight = 2.3;
snack[0]->calories = 350;

Replace all of the -> operators with the . operator.

snack is a pointer, but using the subscript operator dereferences the
pointer. Thus snack[0] is of type candybar, not pointer to candybar.
--
John Carson

=============== =============== =============== =====
This Programme has a litte problem.First using string class ,you must use
headfile <string>
The other hand,I suggest you'd better use form like string a (" ") to
initiate variable of string class.

[Howard]

"Karl Heinz Buchegger" <kb******@gasca d.at> wrote in message > So in short:

Even if snack is a pointer, snack[0] is no longer of pointer type, but is
already
the object in the array. Therefore you don't use ->, but simply access the
component
directly with a '.' operation.

Really? Wow...that's just strange. I guess I've never used an array of
pointers like that, because I sure didn't know this. (I tend to use vectors
nowadays for most dynamic array uses.)

So, suppose I actually _need_ a pointer from that array. How do I get it?

Like this?

candybar* pBar = &(snack[i]); // Like this?

That sure seems odd.

Somehow, I thought I'd used this kind of notation before somewhere:

(snack[i])->name = "Snickers";

Would that work?

-Howard

[John Carson]

"Howard" <al*****@hotmai l.com> wrote in message
news:CN******** ***********@bgt nsc04-news.ops.worldn et.att.net

"Karl Heinz Buchegger" <kb******@gasca d.at> wrote in message > So in
short:

Even if snack is a pointer, snack[0] is no longer of pointer type,
but is already
the object in the array. Therefore you don't use ->, but simply
access the component
directly with a '.' operation.

Really? Wow...that's just strange. I guess I've never used an array
of pointers like that, because I sure didn't know this. (I tend to
use vectors nowadays for most dynamic array uses.)

There is no array of pointers, just a pointer to an dynamically allocated
array. The pointer name functions just the same as the name of a statically
allocated array in terms of accessing elements of the array.
So, suppose I actually _need_ a pointer from that array. How do I
get it?
Like this?

candybar* pBar = &(snack[i]); // Like this?
snack is a pointer, as is snack + i, so you can write:

candybar* pBar = snack + i;
That sure seems odd.
Not to me. Using a subscript on a pointer is equivalent to dereferencing
the pointer. To make the point clear, consider:

int x = 1;
int *ptr = &x;

Then ptr[0] is the same as *ptr, i.e., it is equivalent to x.
Somehow, I thought I'd used this kind of notation before somewhere:

(snack[i])->name = "Snickers";

Would that work?

No.
--
John Carson

[Howard]

"John Carson" <jc************ ****@netspace.n et.au> wrote in message
news:42******** *************** @un-2park-reader-02.sydney.pipen etworks.com.au. ..

"Howard" <al*****@hotmai l.com> wrote in message
news:CN******** ***********@bgt nsc04-news.ops.worldn et.att.net

"Karl Heinz Buchegger" <kb******@gasca d.at> wrote in message > So in
short:

There is no array of pointers, just a pointer to an dynamically allocated
array.

D'oh! I got confused. Somehow, I was thinking that he'd allocated an array
of pointers. But, that would have been:

candybar** snack = new (candybar*)[10];

My bad. I'll go slink off into a corner and hide my face now.

-Howard

[Benny Andersen]

On Fri, 01 Apr 2005 22:27:10 -0800, Steven Taylor
<ne******@super jacent.net> wrote:

Besides, what is mentioned, you have to use:
delete[] , when deallocating what was allocated using new objects[]

Regards
Benny Andersen