[C ++ pass-by-value]: can the content of the original variables get modified by the called function?

Question

I have always assumed that passing variables with [pass-by-value] in [c++], makes a copy of them, and so the function receiving these copies can not change the original variable's content.

I guess it is because when argument is passed by value, [copy-constructor] is called, and if it is not overridden by the programmer, the default [copy-constructor] does [shallow-copy] instead of [deep-copy]!

So my question is that why they called it pass-by-value, while in the case of having pointers in the class, the function has access to the reference of that pointer and can damage it. Also is this conclusion correct? "Whenever have pointers in the class, and you might pass an object of that class by value as a function argument, define the copy constructor!"

If this error is famous, does anyone know the name or phrase to this problem?

Below is my code that resulted in a modified list. Objects of this class contain an int member variable as well as a pointer to a list of nodes member variable.

class ComplicatedObject{  
  public:
    ComplicatedObject();
    ~ComplicatedObject();

    //ComplicatedObject(const ComplicatedObject& c);  // copy construtor
    //ComplicatedObject& operator=(const ComplicatedObject& c); // =operator

    int int_member_varialbe_;  // an int member variable
    void addToList(int d);
    void printList();

  private:
    struct node{
      int data;
      node* next;
    };
    node* head_;  // points to the beginning of a list (linkedlist of int)
};

The below code prints 2. And then prints 2 3 !

Test.cpp:

void myfunction(ComplicatedObject obj){
  obj.addToList(3);
  obj.int_member_variable_ = 5;
}

int main(void){
  ComplicatedObject c_object;
  c_object.addToList(2);
  c_object.printList();  //prints 2 

  cout << "int member variable befor passing:";
  cout << c-object.int_member_variable_ << endl;  //prints 6 (a default value)

  myfunction(c_object); //pass-by-value

  cout << "int member variable after passing:";
  cout << c-object.int_member_variable_ << endl;  // prints 6 (not 5)

  c_object.printList(); // prints 2 3 ! List get changed!

  return 0;
}

The list gets changed because you are doing a shallow copy of the object. — Captain Obvlious, Jul 03 '13 at 19:26
Pointers have a value, and by default that value is copied. This value is the address of another value. You could write a `value_ptr` smart-pointer class that by default copies-by-duplication (be it clone, copy construction, or what have you). As to why this happens by default? Well, default `operator=` on `struct` and `class` behavior was determined before C++ split from C. — Yakk - Adam Nevraumont, Jul 03 '13 at 19:30
Imagine worst, what will happen if you delete the first pointer and after that try to access second one! Btw the golden rule is to almost never use raw pointers. — CyberGuy, Jul 03 '13 at 19:31

score 2 · Answer 1 · answered Jul 03 '13 at 19:26

You are correct. A shallow copy (which occurs if your class lacks a copy constructor) copies the pointers in your class, but not what they point to. You must define a copy constructor which performs a deep copy if you'd like to be able to properly pass a non-const object by value.

score 1 · Answer 2 · answered Jul 07 '13 at 05:43

I have always assumed that passing variables with [pass-by-value] in [c++], makes a copy of them, and so the function receiving these copies can not change the original variable's content.

Yes, that's true.

I guess it is because when argument is passed by value, [copy-constructor] is called, and if it is not overridden by the programmer, the default [copy-constructor] does [shallow-copy] instead of [deep-copy]!

A "copy" is a "shallow copy". A "copy" copies all the contents of the objects. And the contents are the fields. And if the field is a pointer, then the content is the pointer, the address. That's it.

So my question is that why they called it pass-by-value, while in the case of having pointers in the class, the function has access to the reference of that pointer and can damage it.

So? The things that those pointers point to are not part of the object. So it's irrelevant. The pointer is what's part of the object.

score 0 · Answer 3 · answered Jul 03 '13 at 20:13

So my question is that why they called it pass-by-value, while in the case of having pointers in the class, the function has access to the reference of that pointer and can damage it.

Let's consider the following situation:

int func(int b, int* c)
{
  /* some stuff */
}

for the first parameter, b, it is a "ByVal" delivery, so what the compiler usually does is

Create sizeof(int) space for b
Copy b to the space
The func can modify the local copy of b without affecting original b

This is necessary, since if it is automatically delivered "ByRef", compiler will meet trouble when non L-value is given, i.e. 1, as you cannot get the reference from a constant value.

Now let's see what happens to c. Again, it has a type of int*, not a int&, so it is treated the same way like int b.

However when c is a int* pointer, it does not store any other data except the address to some object with type int. So indirect access via the pointer is permitted and there is only copy of the address but not the value it points to,thus can be damaged.

[C ++ pass-by-value]: can the content of the original variables get modified by the called function?

3 Answers3