C++ new() operator

June 28, 2012 at 7:17 pm | Posted in C++, Programming | Leave a comment
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I was working on adding some features to a software written as a mixture of C and C++ and I came to realize many C++ programmers write C in C++. Take a look at this:

#include <iostream>
#include <new>

int main()
{
  char * t = new char[1];

  if(!t)
    {
      printf("Memory Exhausted\n");
      exit(EXIT_FAILURE);
    }

  t[0] = 'A';

  printf("t[0] = %c\n", t[0]);
  return 0;
}

Everything looks fine and g++ (on RHEL 5) even compiles/runs with flags (-ansi -pedantic -Wall -Wextra) and without any warning/error. Program just seems to be technically correct (but it isn’t) and prints the value inside array as expected.

Now let us take a more basic look at this. Why we are using printf() in a C++ program when C++ has its own standard library? The program we have written is actually C in C++. It is like using a powerful mechanism and throwing all the powerful tools/methods away because you don’t want to know about them. Many industrial programmers don’t want to know about it because life goes on fine without knowing about it. It does not cause any immediate problems. Hence Why bother? Wrong reasons. Rewrite it the C++ way:

#include <iostream>
#include <new>

int main()
{
  char * t = new char[1];

  if(!t)
    {
      std::cerr << "Memory Exhausted" << std::endl;
      exit(EXIT_FAILURE);
    }

  t[0] = 'A';
  std::cerr << "t[0] = " &lt;&lt; t[0] << std::endl;

  return 0;
}

It is still not C++. What we call it now is a mixture of C and C++ provided the programmer does not know about C++ much. He was basically a C programmer and was either forced to use C++ compiler or was too lazy to know C++. Again it compiles/runs fine as it looks technically correct. check out new operator in section 18.6.1 of current C++ standard (you can check out draft of the standard, just google for N3337.pdf). new does not return NULL, hence the check is wrong. new throws bad_alloc() if it can’t allocate memory. Hence the correct version looks like this:

#include <iostream>
#include <new>

int main()
{
  char * t;

  try { t = new char[1]; }
  catch(bad_alloc)
    {
      std::cerr << "Memory Exhausted" << std::endl;
      exit(EXIT_FAILURE);
    }

  return 0;
}

This will never compile. Why? Because you did not mention where bad_alloc is coming from. It is in standard library, hence you need to use std::bad_alloc. If you think you simply dump the whole standard namespace by using using namespace std in there, then you really need to read C++ FAQs before writing C++ programs any further. You need to think why I am writing std::cout instead of usual cout? Let us try one more time:

#include <iostream>
#include <new>

int main()
{
  char * t;

  try { t = new char[1]; }
  catch(std::bad_alloc)
    {
      std::cerr << "Memory Exhausted" << std::endl;
      exit(EXIT_FAILURE);
    }

  t[0] = 'A';
  std::cout << "t[0] = " << t[0] << std::endl;

  return 0;
}

This seems like pure C++. Yes, it is. Sure?

We still have one general programming problem remaining to be solved.

When a game adjusts your resolution to 640×480 and resets it back to normal (say 1280×1024), it will call destructor on 640×480 in resetting but it will not do so because exit() will not call the destructor, you will get stuck in a low resolution screen. It happens when some games crash without throwing an exception and rolling back the stack. So what do you do? Let us try one last time to write pure C++ code 🙂

#include <iostream>
#include <new>

int main()
{
  char * t;

  try { t = new char[1]; }
  catch(std::bad_alloc)
    {
      std::cerr << "Memory Exhausted" << std::endl;
      return EXIT_FAILURE;
    }

  t[0] = 'A';
  std::cout << "t[0] = " << t[0] << std::endl;

  return 0;
}

NOTE: There is still a way if you want to check for NULL (instead of catch()ing an exception). You can use the nothrow version of new operator which returns a NULL pointer instead of throwing an exception but you have to mention it explicitly.


Copyright © 2012 Arnuld Uttre, Village – Patti, P.O – Manakpur, Tehsil – Nangal, Distt. – Ropar, Punjab (INDIA)

Verbatim copying and distribution of this entire article are permitted worldwide, without royalty, in any medium, provided this notice, and the copyright notice, are preserved.

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