C ++中的自动指针（auto_ptr）

Question

I am trying to figure out what this piece of code prints but I couldn't output it for some reason, it gave me an error: "1 [main] Q1c 5752 open_stackdumpfile: Dumping stack trace to Q1c.exe.stackdump".

double *dp=new double(1.2);
auto_ptr <double> autodp1(dp);
auto_ptr <double> autodp2=autodp1;
cout<<*autodp1<<endl;

I just want to know what it will print, if it even prints.

Notice: this question was in past exam paper, just for revision.

Answer 1

The code *autodp1 is effectively a dereferencing of a null pointer. Therefore the code exhibits undefined behavior.

You first construct autodp1 to point to the newly allocated double . But then the constructor of autodp2 takes the owned memory for itself and sets autodp1 to null.

Answer 2

That's becouse the operator assignement of auto_ptr takes's the ownership (move) the pointer

Take a read on Wiki, it's quite good general explanation:

http://en.wikipedia.org/wiki/Smart_pointer

"The copy constructor and assignment operators of std::auto_ptr do not actually copy the stored pointer. Instead, they transfer it, leaving the previous std::auto_ptr object empty. This was one way to implement strict ownership, so that only one auto_ptr object could own the pointer at any given time. This means that auto_ptr should not be used where copy semantics are needed."