Solved

Pointers in C++

Posted on 2011-03-01
6
214 Views
Last Modified: 2012-05-11
Hello experts.

void aaa(bbb* aVar)
{
      bbb myVar;
      
      myVar = aVar[1]; //compiles
      //myVar = *aVar[1]; //does not compile

}

This is my thinking process:

a. bbb * aVar

aVar is a pointer to address of bbb.

We can treat aVar as an array where
aVar[0] points to first address of bbb
aVar[1] points to second address of bbb, and so on.

So...

bbb = aVar[1] should make bbb equal to address of second
element of aVar

bbb = *aVar[1] should make bbb equal to value of second
element of aVar


Where do I get it wrong?

Thank you

panJames


0
Comment
Question by:panJames
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 4
6 Comments
 
LVL 75

Accepted Solution

by:
käµfm³d   👽 earned 500 total points
ID: 35006367
>>  aVar is a pointer to address of bbb.

aVar is a pointer to the location of an object of type bbb. The indexing operation ( [n] ) is equivalent to saying

    *(somePtr + n)

What you are trying to do with this line:

    myVar = *aVar[1];

is dereference whatever is found in the first slot of the array aVar.
0
 
LVL 75

Expert Comment

by:käµfm³d 👽
ID: 35006395
>>  dereference whatever is found in the first slot of the array aVar.

Using the aforementioned equivalence, what you attempting to do is:

    **(aVar + 1)
0
 

Author Comment

by:panJames
ID: 35006411
//
What you are trying to do with this line:

    myVar = *aVar[1];

is dereference whatever is found in the first slot of the array aVar.
//


why is it illegal?

panJames
0
Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
LVL 6

Expert Comment

by:Mistralol
ID: 35006934

Thats kinda of true but what you are really doing is this

**(avar + sizeof(bbb));


void aaa(bbb* aVar)
{
      bbb myVar;
     
      myVar = aVar[1]; //compiles
      //myVar = *aVar[1]; //does not compile

}

The first works because myVar is taking a copy of the data in aVar[1] from the array slot 1.
So its the same as memcpy(&myVar, &aVar[1], sizeof(myVar));







0
 
LVL 75

Expert Comment

by:käµfm³d 👽
ID: 35007123
My C-Fu is not that strong, but IIRC, unless aVar is a pointer to an array of pointers, array of arrays, or a pointer to a pointer, you will get an error.
0
 
LVL 75

Expert Comment

by:käµfm³d 👽
ID: 35007128
*with the above syntax  = )
0

Featured Post

Free Tool: Path Explorer

An intuitive utility to help find the CSS path to UI elements on a webpage. These paths are used frequently in a variety of front-end development and QA automation tasks.

One of a set of tools we're offering as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Often, when implementing a feature, you won't know how certain events should be handled at the point where they occur and you'd rather defer to the user of your function or class. For example, a XML parser will extract a tag from the source code, wh…
Many modern programming languages support the concept of a property -- a class member that combines characteristics of both a data member and a method.  These are sometimes called "smart fields" because you can add logic that is applied automaticall…
The goal of the video will be to teach the user the difference and consequence of passing data by value vs passing data by reference in C++. An example of passing data by value as well as an example of passing data by reference will be be given. Bot…
The viewer will be introduced to the technique of using vectors in C++. The video will cover how to define a vector, store values in the vector and retrieve data from the values stored in the vector.

623 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question