Need to know why destructor is calling 04 times?

Question

There was a problem in our last class.(http://ideone.com/oaNbfd . i could solve it. but the problem was with the output. Here constructor should call 02 times and the destructor should call 02 times. but destructor was calling 04 times .Now ques is why it is 04 times???output was like that:
constructing1
constructing2
destructing1
destructing1
destructing1
destructing1

input file link : http://ideone.com/oaNbfd

Vulpes · Answer

This is an example of 'implicit' copy constructors which can be a problem in C++ code. The code to concentrate on is this function: myclass func(myclass ob) { return ob; } and this line which calls it: b = func(a); When func is called a bitwise copy is created of the myclass object 'a' and assigned to the parameter 'ob'. When this function returns this copy goes out of scope and the destructor is called. Also this line: return ob; creates a second bitwise copy of 'a' which is what is actually returned. When this object is assigned to 'b' there are two consequences: 1. The second bitwise copy goes put of scope and the destructor is called a second time. 2. The object 'b' has its 'id' field of 2 overwritten with 'a's value of 1. When the main() method ends the destructors are called on the objects 'a' and 'b' which both now have an 'id' field of 1. So the destructor is called 4 times altogether with an id of 1 in each case. Notice that since myclass doesn't have a copy constructor a default one is generated in the background. The 'ordinary' constructors are not called and so you're not notified that extra objects have been created.