Count the number of objects created statically and dynamically in C++

Question

I have written below program to count the number of static and dynamically created object of the class.

#include <iostream>
class Test 
{
   public:
    Test()
    {
      stackCount++;
    }
    
    void* operator new(size_t objSize)
    {
       void* ptr = malloc(objSize);
       heapCount++;
       return ptr;
    }
    
    static void Display()
    {
      std::cout << "stack object count : " << stackCount - heapCount << ", Heap object Count :" <<  
      heapCount << std::endl;
    }
    
   private:
    static int stackCount;
    static int heapCount;
};
    
int Test::stackCount = 0;
int Test::heapCount = 0;
     
int main() 
{
    Test obj1;
    Test obj2;
    Test obj3;
       
    Test *ptr = new Test();
        
    Test::Display();
        
    return 0;
}

Program output : stack object count : 3, Heap object Count :1

is there any better way to maintain the count of static and dynamic object.?

Answer 1

malloc is used but never defined. You probably want std::malloc, from the <cstdlib> header.

Prefer pre-increment (++i) to post-increment, if you're not using the result. For simple arithmetic types, decent compilers will optimise away the copy, but that's not possibly for more complex class types, so cultivate a good habit that can save some performance one day.

On the subject of small performance wins, don't use std::endl when there isn't a need to flush the stream (and prefer a separate std::flush even then, to keep the intention clear).

I recommend changing how you break long << lines, to make it clearer what's a continuation line even when you can't see the whole line:

      std::cout << "stack object count : " << stackCount - heapCount
                << ", Heap object Count :" << heapCount << '\n';

See how the indentation and use of << at the start help show that this is a single (logical) line?

You have a memory leak, because you never delete ptr. However, we need to implement a matching operator delete() for that to work properly.