Are both method declarations "pass by reference"?

What is the difference between these two ways of writing a function in C++? Are they both "pass by reference"? By "pass by reference", I mean that th开发者_开发百科e function has the ability to alter the original object (unless there is another definition, but that is my intention).

From my understanding, when you call f1, you pass in a "synonym" of the original object. When you call f2, you are passing in a pointer to the object. With the f2 call, does a new Object* get created whereas in with f1 call, nothing does?

f1 (Object& obj) {}
f2 (Object* obj) {}

Thanks!

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Your understanding is correct.

Technically, f2 is pass by value, with the value being a pointer, but you can use that copy of the pointer to modify the pointed-to object.

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They are both pass-by-reference in a general sense, but that phrase will not normally be used to describe the pointer version because of possible confusion with C++ reference types.

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What is the difference between these two ways of writing a function in C++? f1 is passing an object by reference, f2 is passing a pointer by value. Both have the ability to modify the object f1 directly through the reference and f2 by dereferencing the pointer.

With the f2 call, does a new Object* get created whereas in with f1 call, nothing does? f2 could allocate a new object at that address or use an already allocated object at that address depending. However passing a pointer by value will not call new it will simply be a copy of the pointer which may or may not point to a valid allocated object.

Check out the wiki link posted as a comment by pst.

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f1 (Object& obj) {}

When you call f1, for example f1(o1):

• obj becomes the other name for o1 (alias), that's it. So o1 is just alias for obj; they have the same address (you can check that &obj==&o1 is true)

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f2 (Object* obj) {}

When you call f2,for example f1(&o1):

• obj is created and initialized with &o1; that is similar like:

 f2(){
    Object* obj=&o1; // just for understanding what happens when you call f2(&o1)
    ...

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A pointer can be null where a reference can not.

See:

What are the differences between a pointer variable and a reference variable in C++?