unique_ptr behavior in c++

Question

Odd behavior in object creation using std::unique_ptr. Here are two example :

#include<iostream> 
#include<memory> 
using namespace std; 

class A 
{ 
public: 
    A() {
        throw "EOF";
    }
  void show() 
  { 
    cout<<"A::show()"<<endl; 
  } 
}; 

int main() 
{ 
    try {
      unique_ptr<A> p1 = make_unique<A>(); 
      p1 -> show(); 
    } catch(...) {
        cout << "Exception" << endl;
    }     
  return 0; 
}

Output

Exception

This above output looks obvious. But, the below code's output is Odd.

// C++ program to illustrate the use of unique_ptr 
#include<iostream> 
#include<memory> 
using namespace std; 

class A 
{ 
public: 
    A() {
        throw "EOF";
    }
  void show() 
  { 
    cout<<"A::show()"<<endl; 
  } 
}; 

int main() 
{ 
    try {
      unique_ptr<A> p1; 
      p1 -> show(); 
    } catch(...) {
        cout << "Exception" << endl;
    }
  return 0; 
}

Output

A::show()

These example are compiled with c++14 compiler. Is the above output expected behavior?

That's undefined behaviour. – HolyBlackCat Oct 08 '18 at 18:06 — HolyBlackCat, Oct 08 '18 at 18:06

score 2 · Answer 1 · answered Oct 08 '18 at 18:12

Your second example does not actually create an A object. It merely creates a unique_ptr to an A that is not initialized. Dereferencing such a pointer is Undefined Behaviour and the compiler is allowed to do whatever it wants, but you cannot rely on any specific behaviour. Example two is broken code.

score 1 · Answer 2 · answered Oct 08 '18 at 18:09

Is the above output expected behavior?

Yes, and no. There is no correct behaviour, because the behaviour of indirecting an empty smart pointer is undefined, just like indirecting a null pointer is. So, you shouldn't expect anything in particular, but nothing is unexpected either.

unique_ptr behavior in c++

2 Answers2