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3 cons and 3 dest for each function
Test1
Base()
Base(const Base&)
Base(const Base&)
~Base()
~Base()
~Base()

Test2
Base()
Base(const Base&)
Base(const Base&)
~Base()
~Base()
~Base()

3 ctor 3 dtor for each function

can anybody explain how r 3 constructors getting called in test1 i can see only 2 one being default constructor and another copy constructor which copies by value

3 Constructor and 3 destructor.
@destiny read the "temporary" topic of C++.

>> can anybody explain how r 3 constructors getting called in test1 i can see only 2 one being default constructor and another copy constructor which copies by value

A temporary object is created when return b; is called in func() - the return type of func is Base.

Please explain detail..why 3 constructor called for test1 and test2 each

Test1 :
1) default will be called at Base a
2) Copy constructor will be called when a is passed to func
3) Copy constructor will be called when b is returned from func

Test2 :
1) default will be called at Base a
2) Copy constructor will be called when a is passed to func
3) Copy constructor will be called when b is returned from func

Test1
Step-1 Default constructor will be called at Base a;
Step-2 Copy constructor will be called when passing a as formal parameter i.e func(a).
Step-3 Copy constructor will be called for return object.
Step-4 Destructor will be called for actual parameter of func.
Step-5 Now this return obj will act as nameless obj in test1() and will get destroy after line func(a) calling destructor one more time.
Step-6 Finally Destructor will be called for a.

You can see same for test2();

Not sure why test2 has only 4 constructor ...

For Base b = func(a), when the 3rd constructor is called to generated the temporary return obj, the temporary obj just get a "new name" - b - so no new copy constructor needed... right?

This is a compilaton error. Because,test1()/test2()doesn't exists.
And there is redfination of test().

compile and run. Then you will see for yourself. No 4 constructors for test2.

If you want to sound real smart in your interview, you can say that the answer to 3 depends on whether your compiler employs RVO (Return Value Optimization), which it likely thus. In this case, of the Unnamed variety.

It would be 4 otherwise (1 default and 3 copy constructors). But the compiler sees the following

Base b = func(a);
Base func(Base c){return c;}

and rewrites it as

Base b(c);

Simply put, the compiler will construct the temporary inside the space saved for object b.

i agree with above comment...most of the compilers do return value optimization, so that might affect the no of constructor and destructor calls

there is no function test1 and test2

same number of function calls for test1 and test2
Base()
Base(const Base&)
Base(const Base&)
~Base()
~Base()
~Base()

3 sets of constructors will be called for both cases, and in second case the newly created object is b, whether for first case newly created object is unnamed and is not stored in a variable.