A question regarding the duality of pointers and arrays and another on structures.

[Squignibbler]

Hi all,

I have a question regarding the C++ programming language
regarding the nature of the relationship between pointers and arrays.

If the statement MyArray[x] is functionally identical to
*(MyArray+x), what statement is functionally identical to
MyArray[2,x]?

I ask this question because when I create a dynamic array with one
dynamic dimension e.g p_MyArray = new int [2,x] how do I access the
individual elements of the array using the pointer p_MyArray?
On a similar note, If I have created a dynamicly sized array of
structures, how do I retrieve the address of the member variable in any
given structure in the array?

I am aware that X = p_StructreArray->Member_variabl e will
retrive the value of MemberVariable through pointer p_StructureArra y,
but how do I get the address?!

Thanks for any help or insight you can offer a beffuddled mind!

[Phlip]

Squignibbler wrote:

I have a question regarding the C++ programming language
regarding the nature of the relationship between pointers and arrays.

If the statement MyArray[x] is functionally identical to
*(MyArray+x), what statement is functionally identical to
MyArray[2,x]?
Allow me to be the first of many.

The expression 2,x returns x. The 2 is thrown away. So you have a flat
array, not a two-dimensional array.

Get that with MyArray[2][x]. Now the array aggregate is recursively defined.

And pointers and arrays are not themselves dual. Accessing them with [] is.
On a similar note, If I have created a dynamicly sized array of
structures, how do I retrieve the address of the member variable in any
given structure in the array?

I am aware that X = p_StructreArray->Member_variabl e will
retrive the value of MemberVariable through pointer p_StructureArra y,
but how do I get the address?!

&p_StructreArra y[x].Member_variabl e

The [] and . precede the &, so it evaluates last, and returns the address of
one of the Member_variable s.

--
Phlip
http://industrialxp.org/community/bi...UserInterfaces

[Squignibbler]

Thankyou very much indeed for your accurate, and speedy advice. You
have helped greatly in my understanding of the situation. Thanks again,
SquigNibbler

[Squignibbler]

Please I must ask for some further help:

Umm....okay now im really confused!! if I use

cout << &p_StructreArra y[x].Member_var*iab le I see an address as
one would expect...

however if I assign a pointer the value and print that i get the value
of the member variable!!

char *p_AVariable;
.........
........

p_AVariable = &p_StructreArra y[x].Member_variabl e;
cout << p_AVariable; //this display the meber variable not the
address...why?!

IF anyone could shed an light on my ignorance i would be very gratefull!

[Phlip]

Squignibbler wrote:

cout << &p_StructreArra y[x].Member_var*iab le I see an address as
one would expect...

What type is Member_variable ?

<< depends on one overloaded operator<< for each type that can stream. If
the input isn't char*, the operator that streams a pointer's address will
compile in.

When you assign to a char * and use that, you might be shifting the type to
one that << interprets as a string.

--
Phlip
http://industrialxp.org/community/bi...UserInterfaces

[Squignibbler]

Quote: Philip wrote..
"What type is Member_variable ? "

It is a array of 30 chatacters, used to hold a string read from a file.
Quote: Philip wrote..
"When you assign to a char * and use that, you might be shifting the
type to
one that << interprets as a string."

Thus it makes alot of sense, although the string is appearing when try
and pass the address aquired by the
"&p_StructreArr ay[x].Member_variabl e" statement to a variable or use it
as the return value of a function. Please could you suggest away round
this problem, so i can store the address's retrieved using the
"&p_StructreArr ay[x].Member_var*iab le" method?

Thanks for your help, your helping more than you can imagine. :D :D
:D

[Mike Wahler]

<Sq**********@h otmail.com> wrote in message
news:11******** **************@ l41g2000cwc.goo glegroups.com.. .

Hi all,

I have a question regarding the C++ programming language
regarding the nature of the relationship between pointers and arrays.

If the statement MyArray[x] is functionally identical to
*(MyArray+x), what statement is functionally identical to
MyArray[2,x]?
*(MyArray + x)
Read about the comma operator.
I ask this question because when I create a dynamic array with one
dynamic dimension e.g p_MyArray = new int [2,x]
That array has a single 'dimension', that is, 'x'.
See the FAQ for how to allocate 'multidimension al'
arrays with 'new':
http://www.parashift.com/c++-faq-lit...html#faq-16.16

However, I'd recommend using containers instead (e.g.
std::vector<std ::vector>. Then all the memory management
is handled for you automatically.
how do I access the
individual elements of the array using the pointer p_MyArray?
On a similar note, If I have created a dynamicly sized array of
structures, how do I retrieve the address of the member variable in any
given structure in the array?

I am aware that X = p_StructreArray->Member_variabl e will
retrive the value of MemberVariable through pointer p_StructureArra y,
but how do I get the address?!
Same way you get the address of any object, use the address operator.
&p_StructreArra y->Member_variabl e

Thanks for any help or insight you can offer a beffuddled mind!

Which C++ book(s) are you reading that don't explain these
issues?

-Mike

[Squignibbler]

C++ from the ground up.. .. its an older book from 1994... it does talk
about the & operator, but I have just had no success with usign it in
this one program i wrote!

int * X;

X = &p_StructreArra y->Member_variabl e; //is a char[30] array

cout << X; //this displays the word in Member_variable :-(

cout << &p_StructreArra y->Member_variabl e; //this displays the

address :-)

now philip mentioned the fact that my compiler may be performing a
substitution because X is a pointer to an array of characters...
OR is there any other valid reason why this may occur!?

Quote "However, I'd recommend using containers instead (e.g.
std::vector<std ::vector>. Then all the memory management
is handled for you automatically."

Thanks!! This might be just what i ned to learn next. I must say it is
amazing that people are prepared to take time to help others :D