Casting to pointer - managed or unmanaged?

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Hi,

Let's say I have the following code:

C++

consoleHelper.write_status("Error: " + (char*)gai_strerror(err_code));

There's a cast there to a char* and there's no delete called on it. Is this memory somehow automatically managed, or would you have to do something like the following?

C++

char* cast_ch = (char*)gai_strerror(err_code);
consoleHelper.write_status("Error: " + cast_ch);

delete cast_ch;

Even if the memory used is tiny, I'd like to prevent all memory leaks wherever possible to ensure the application can theoretically run forever with no issues.

What I have tried:

Nothing, just asking for advice

Posted 6-Jul-18 1:37am

User 13204940

Updated 6-Jul-18 3:27am

v5

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Richard MacCutchan 6-Jul-18 9:23am

No you cannot delete it unless the documentation for that function explicitly tells you to.

2 solutions

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CPallini · Accepted Answer · 2018-07-06T02:17:00

Solution 1

You should NOT release memory (it is not explicitely required in the documentation and, moreover, you don't know how the library actually allocates it).

Please note: if there is a leak, that's a (unlikely) library's author mistake (and probably harmless). On the other hand, calling delete on the pointer would be dangerous and entirely of your responsibility. :-)

Posted 6-Jul-18 2:17am

CPallini

Updated 6-Jul-18 2:21am

v2

Comments

[no name] 6-Jul-18 8:21am

Which library are you talking about? Which documentation? The code behind the cast cannot tell when it's finished being used, so it can't possibly free the memory itself, surely?

CPallini 6-Jul-18 8:35am

The library implementing the gai_strerror function. The documentation of such a library. If you don't know how memory is allocated (and what C runtime did allocate it) the you must NOT release it.

[no name] 6-Jul-18 8:43am

All of the code apart from gai_strerror is my own.

CPallini 6-Jul-18 9:50am

The string comes just from gai_strerror.
Let's face it: trying to release memory on such a pointer is a plain mistake.

[no name] 6-Jul-18 10:58am

But I want to understand the reason *why*.

CPallini 6-Jul-18 16:41pm

Then you have to study. A starting point could be this article:
https://blogs.msdn.microsoft.com/oldnewthing/20060915-04/?p=29723

Richard MacCutchan · Accepted Answer · 2018-07-06T03:27:00

Solution 2

The documentation (which you should always check first) : gai_strerror[^] tells you that the return is a pointer to a character constant, so you should not do anything with it.

Posted 6-Jul-18 3:27am

Richard MacCutchan

Comments

[no name] 6-Jul-18 10:54am

Why exactly does it being a const char* mean you shouldn't do anything with it? I'm trying to understand the reasons why these things are what they are.

Richard MacCutchan 6-Jul-18 11:47am

Because the thing being pointed at is a constant, that is to say, it cannot be modified. And you cannot create a constant by using new or malloc, since dynamically created memory space can always be modified.

[no name] 6-Jul-18 11:50am

Ok so const pointers never need to be deleted in any scenario?

Richard MacCutchan 6-Jul-18 11:54am

That's a good rule of thumb. Although the use of const can sometimes cause confusion, see The C++ 'const' Declaration: Why & How[^].

Richard MacCutchan 6-Jul-18 11:58am

And another issue to be careful of. You should not cast that pointer to a simple char*, as you are casting away its const'ness, thus leaving it open to misuse.

CPallini 6-Jul-18 16:59pm

That's not completely true. An (admittedly contrived) example:

#include <stdlib.h>
#include <stdio.h>

static char * p = (char *)0;

void init_library();
const char * get_const_string();
void terminate_library();


void init_library()
{
  if ( ! p )
  {
    p = (char * ) malloc(6);
    p[0]='h'; p[1]='e'; p[2] = 'l'; p[3] = 'l'; p[4] = 'o'; p[5] = '\0';
  }
}

void terminate_library()
{
  if ( p ) free (p);
}

const char * get_const_string()
{
  return p;
}

Richard MacCutchan 7-Jul-18 2:55am

Well you can contrive anything if you try hard enough.

CPallini 7-Jul-18 3:22am

Indeed. However the code sample is contrived, not the point. We usually think a const char * pointer addresses read-only memory but that is not always true.

Richard MacCutchan 7-Jul-18 3:29am

I agree completely, but I was deliberately avoiding this point in order to try and keep things simple.

Richard MacCutchan 7-Jul-18 3:36am

And of course the following will be caught by the compiler:

    const char *s = get_const_string();
    s[0] = 'B';

// and ...
    const char *s = get_const_string();
    free (s);

CPallini 7-Jul-18 3:56am

However those do not change the matter: you cannot assume how the string is actually allocated.

Richard MacCutchan 7-Jul-18 4:06am

No, but the whole point is that the const qualifier is to help you avoid careless mistakes. If you misuse it then you are responsible for any problems.

Rick York 6-Jul-18 17:10pm

To state, "you shouldn't do anything with it" is entirely incorrect. You can print it, copy it, and many other things. The only thing you can not do to it is change it because it is "const." Freeing or deleting would amount to changing it so those are not allowed.

Also, as stated elsewhere, it should not be cast to a non-constant value either. There are several reasons for that but one of them is most string class accept const character strings but not non-const ones so that is not an appropriate cast in your case. I mentioned this because in your code you have this, "consoleHelper.write_status("Error: " + cast_ch")", which looks like you are using the + operator from a string class to append strings and usually that won't work unless both arguments are "const char *" or string objects.

Richard MacCutchan 7-Jul-18 2:57am

I was responding in the context of whether it is OK to use free or delete on such a pointer. The other uses go without saying. And it was me who said not to cast away const.