class object equals class pointer object with templates

Question

I don't know a question for my C++ course.

template <typename T> class Array {...};
void main(int argc, char* argv[]) {

Array<int> *a1 = new Array<int>();
Array<int> a2();
a2 = *a1; // <--  
}

What will happen at the noted line (<--)? a) a2 copy constructor is called. b) The address of a2 becomes equal to the address of pointer a1. c) copy assignment operator of a2 is called d) compilation error

Justify your answer.

You know you could write the code run it and find out. – NathanOliver Jul 08 '15 at 15:28 — NathanOliver, Jul 08 '15 at 15:28
This question does not show any research effort. – Drew Dormann Jul 08 '15 at 15:31 — Drew Dormann, Jul 08 '15 at 15:31

score 1 · Answer 1 · answered Jul 08 '15 at 15:31

1

d) compilation error as you have vexing parse:

Array<int> a2(); // Declaration of function which takes no argument and return Array<int>

answered Jul 08 '15 at 15:31

Jarod42

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@TartanLlama My bad. You are correct. It also will generate an error/warning since main is declared as `void` – NathanOliver Jul 08 '15 at 15:41

class object equals class pointer object with templates

1 Answers1