CODEWITHSUNDEEP
c++
MistyD
MistyD

Reputation: 17223

Why is the copy constructor not called in this case - Is my reasoning correct?

Suppose I have an object called Myobj and this object has all the defaults (copy cstr , move cstr, assignment and cpy assignment). Now suppose there is a method as such

void fooA(foo& f){..}  --->A

and then another method as such

void fooB(foo f){..}   --->B

Now when asked why the copy constructor is not called in case of A and is called in-case of B my answer and understanding is as follows. Kindly correct me if its wrong

The reason its not called for case A is because foo& is a reference to foo and not foo therefore the copy constructor of foo is not called. The f in case A is simply an alias to the passed object.

The reason its called in case B is because foo is the object foo and not a reference to the foo.Therefore the copy constructor of foo is called. The cpy cnstr is foo::foo(const foo& f). So incase of the statement 

fooB(someobject);

The equivalent would be on the function as

void fooB(foo f){..} gives  fooB f (someobject); //Cpy constr is called

Kindly let me know if my answer and understanding is correct

Upvotes: 0

Views: 85

Answers (2)

luk32
luk32

Reputation: 16070

Yes, but IMO you are wording is quite strange or maybe you are over-complicating things a bit.

The second signature uses a "pass by value" method. As opposed to the first example, where a "pass by reference" is used. The idea is that you operate on the value of the argument, and not the argument itself.

This means a temporary copy of the passed parameter is made, with the lifetime limited by the scope of the function.

Upvotes: 3

aditsinha
aditsinha

Reputation: 36

Yes that is correct. I think you mean in the first case that it is a reference to f not a reference to foo. Same with in the second case it is that you're passing a copy to f not the f itself.

Upvotes: 0

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