create a dynamic array of object, delete object and free memory in C++

Question

Hello everybody ! I am a beginner in C++ so I don't really have a good experience.

I want some help,

I try to create a dynamic array of object (type ELEMENT), I have a problem when I want to delete an object from array (Error 2157)

This is a part of code :

class ELEMENT
{
private :
     int id_num;
     int nbnr;
     int BI;
public :
: void () ................
: ...............
:  
 };

 ELEMENT *T;   

/* before calling the next fonction, I allocate a dynamic memory space every time for T by
 T = new TAB;  because I don't know the exact size of T, I don't know if it's right like this ?
*/

void eval (int nr, int BS)
{

for (int i=0; i< size; i++)
{if (T [i].BI >= BS)
delete T [i];   // I try to delete the object in position (i) and also free allocated memory
// before I tried with delete [] T; doesn't work !
}

}

and the other question, is there any function to get the current size of T. I tried with SizeOf(T) doesn't give right value.

That's all, thanks for your answers !

see stackoverflow.com/questions/5754864/…

hivert
– hivert

2014-02-21 08:04:49 +00:00
Commented Feb 21, 2014 at 8:04 — hivert
– hivert, Commented Feb 21, 2014 at 8:04

mathematician1975 · Accepted Answer · 2014-02-21 08:02:54Z

2

You need to decide on the size of your array and then allocate dynamically as

T = new ELEMENT[size];

This creates an array of ELEMENT items which you delete once with

delete [] T;

You should not call delete on each element of T. The only time you would need this would be if T contained pointers to dynamically allocated memory (which they do not in your case) so do not do this.

There is no way to get the size of T using a dynamically allocated array. However, if you were to use std::vector instead you would be able to do this and also not have to worry about memory allocation and release at all.

answered Feb 21, 2014 at 8:02

mathematician1975

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3 Comments

hivert Over a year ago

I would not say 'You do not need to call delete on each element of T.' but 'You can't delete the element of T individually'

user3334513 Over a year ago

Thank's for your answer; So with this structure I can't point to an element (specific position) of array and delete it, that's what you mean ?

mathematician1975 Over a year ago

@user3334513 You allocate memory for an array - this is a contiguous block and within this memory region will be the space for each ELEMENT of your array. When you delete the array using delete [] T you are freeing all of the memory - there is no need to do anything else. You should never call delete on the elements of your array unless they are pointers to memory that you have allocated elsewhere. I strongly suggest that once you have understood the correct usage of new and delete that you look at std::vector which will relieve you of the need for memory management of this kind.

songyuanyao · Accepted Answer · 2014-02-21 08:05:24Z

0

Firstly, you can't delete one element from an array and free the memory. You can only delete the while array by delete[].

Secondly, you define T as a pointer of ELEMENT, so sizeof() will return the pointer size but not the array's size. You can remember the size by yourself, or define T as an array such as ELEMENT T[10].

answered Feb 21, 2014 at 8:05

songyuanyao

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Comments

Marko Popovic · Accepted Answer · 2014-02-21 08:14:12Z

-1

With statement ELEMENT *T you have created a pointer to an object of class ELEMENT, but that pointer is not currently initialized ( it is not pointing to any object ). Use

ELEMENT *T = new ELEMENT;

to initialize the pointer to an object.

For the size of an array you can use

sizeof(nameOfYourArray)/sizeof(nameOfYourArray[0])

which calculates the size of an array by dividing total number of bytes with the number of bytes occupied by the first element.

Hope this helps!

answered Feb 21, 2014 at 8:14

Marko Popovic

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1 Comment

Konrad Rudolph Over a year ago

Actually the sizeof code will not work. Try it yourself and think about what it’s supposed to do. And new, while it works, is very bad advice, and doesn’t answer OP’s question.

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