what's wrong?

Joern Dahl-Stamnes -- Jorn.Dahl-Stamnes@fysel.unit.no
Tuesday, January 09, 1996

Hello folks,

Yesterday I wrote some code and had a strange problem. I'm using Visual C++ 1.5
and my code looked like this:

    const char** pStr;

    pStr = new const char* [4];
    .
    .
    .
    delete [] pStr;

The compiler complain about the delete line. It say it cannot convert parameter
1. But if I write:

    delete [] (void*) pStr;

the compiler does not complain.

My question is: WHY? What is wrong with the first delete statement?

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| Joern Yngve Dahl-Stamnes, The University of Trondheim, Norway   |
| e-mail: Jorn.Dahl-Stamnes@fysel.unit.no                         | 
| phone : +73 59 44 12, fax: +73 59 14 41                         |
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---

Dave Brooks -- info@brooksnet.com
Wednesday, January 10, 1996

[Mini-digest: 6 responses]

At 11:17 AM 1/9/96 +0100, Joern Dahl-Stamnes wrote:
>Yesterday I wrote some code and had a strange problem. I'm using Visual C++ 1.5
>and my code looked like this:
>
>    const char** pStr;
>
>    pStr = new const char* [4];
>    .
>    .
>    .
>    delete [] pStr;
>
>The compiler complain about the delete line. It say it cannot convert parameter
>1. But if I write:
>
>    delete [] (void*) pStr;
>
>the compiler does not complain.
>
>My question is: WHY? What is wrong with the first delete statement?

While we often use ** as the address of an array of pointers, strictly
speaking it's not an array type to the compiler.  "delete [] foo"
specifically refers to deleting foo where foo is declared as an array.  You
don't actually need to use the [] operator, but supposedly it optimizes the
delete if foo is an array.  I use it where I can, but haven't noticed any
adverse affects from not using it on non-array types.

I hope this helps!

Best regards,
Dave
**************************************************************************
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-----From: bibhas@hermes (Bibhas Bhattacharya)

Can not delete const pointers.

Bibhas.
bibhas@isgtec.com

-----From: Mario Contestabile 

This code compiles under 4.0.

-----From: Martijn Vels 

I think it has to do with the idea of the delete[] operator.
I'm not sure about this, but here is my 'idea' why the compiler complains.
Think of const char * as object CFoo.
The delete[] operator will call all destructors in the array (. The 
destructor of an object is not const, so you can't delete a const object.
What you have is an array of const objects. The quirk is that allocating 
them whas the first fault which the compiler does not complain about.
Try this:
class CFoo;
const CFoo *p;
p = new const CFoo[2]
The compiler will directly nag you about the alloc.
The only way is:
class CFoo;
const CFoo *p;
p = new  CFoo[2]
...
...
delete[] (CFoo *)p;      /* Not const to us, we own it!!!! */

But, why botter?
The advantage of the const declarator  is mainly for protecting objects you 
pass into functions, or which you return as a result. (CString has therefore 
no operator to autoconvert to char *!).
So the only point I see in using the const declarator is for protecting 
yourself in this specific function.

M.
(vels0020@exact.nl)

-----From: David Stidolph 

>    const char** pStr;
>    pStr = new const char* [4];
>    .
>    delete [] pStr;

The compiler complain about the delete line. It say it cannot convert parameter
1. But if I write:

    delete [] (void*) pStr;


You need to delete each individual string (assuming you assign each pointer with a new):
	for (int i=0; i < 4; i++)
		delete []pStr[i];

In addition, you declared the pointer as const so that might cause your problem.  Try not having the const declaration.

David Stidolph
-----------------------
Some assembly required.

-----From: snkavuri@amoco.com

     This is a C++ question, may be VC++, certainly
not MFC :-)  Coming to your question, the compiler is right.
  In the first delete attempt, you tried deleting a const
  object and it complained.  C++ allows you to coerce  a
  const object to a nonconst one.  You did that the next
  time and it allowed you.

  Surya Kavuri

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