Why is 'this' a pointer and not a reference?

Question

I was reading the answers to this question C++ pros and cons and got this doubt while reading the comments.

programmers frequently find it confusing that "this" is a pointer but not a reference. another confusion is why "hello" is not of type std::string but evaluates to a char const* (pointer) (after array to pointer conversion) – Johannes Schaub - litb Dec 22 '08 at 1:56

That only shows that it doesn't use the same conventions as other (later) languages. – le dorfier Dec 22 '08 at 3:35

I'd call the "this" thing a pretty trivial issue though. And oops, thanks for catching a few errors in my examples of undefined behavior. :) Although I don't understand what info about size has to do with anything in the first one. A pointer is simply not allowed to point outside allocated memory – jalf Dec 22 '08 at 4:18

Is this a constant poiner? – yesraaj Dec 22 '08 at 6:35

this can be constant if the method is const int getFoo() const; <- in the scope of getFoo, "this" is constant, and is therefore readonly. This prevents bugs and provides some level of guarantee to the caller that the object won't change. – Doug T. Dec 22 '08 at 16:42

you can't reassign "this". i.e you cannot do "this = &other;", because this is an rvalue. but this is of type T*, not of type T const . i.e it's a non-constant pointer. if you are in a const method, then it's a pointer to const. T const . but the pointer itself is nonconst – Johannes Schaub - litb Dec 22 '08 at 17:53

think of "this" like this: #define this (this_ + 0) where the compiler creates "this_" as a pointer to the object and makes "this" a keyword. you can't assign "this" because (this_ + 0) is an rvalue. of course that's not how it is (there is no such macro), but it can help understand it – Johannes Schaub - litb Dec 22 '08 at 17:55

My question is, why is this a pointer a not a reference? Any particular reason for making it a pointer?

Some further arguments why this being a reference would make sense:

Consider Item 1 from More Effective C++ : use references when it is guaranteed that we have a valid object i.e. not a NULL (my interpretation).
Furthermore, references are considered safer than pointers (because we can't screw the memory up with a stray pointer).
Thirdly, the syntax for accessing references (.) is a little bit nicer and shorter than accessing pointers (-> or (*)).

@paulm What would this "hack" actually accomplish? Doesn't this always evaluate to true? — iFreilicht
@paulm I don't think that's actually valid C++. Invoking methods on a nullptr to an object results in undefined behavior. — antred
@paulm Maybe it works in some cases, but imagine if the method was virutal. How could a v-table lookup be done with no object? — Jason Creighton
@paulm If you've seen it in production code, abandon ship! That is UB. — Alice
I'm just gonna leave this here... (from MFC's afxwin2.inl): _AFXWIN_INLINE HWND CWnd::GetSafeHwnd() const { return this == NULL ? NULL : m_hWnd; } — Christopher Oicles

Daniel Earwicker Daniel Earwicker · Accepted Answer · 2009-03-14T14:41:27

When the language was first evolving, in early releases with real users, there were no references, only pointers. References were added when operator overloading was added, as it requires references to work consistently.

One of the uses of this is for an object to get a pointer to itself. If it was a reference, we'd have to write &this. On the other hand, when we write an assignment operator we have to return *this, which would look simpler as return this. So if you had a blank slate, you could argue it either way. But C++ evolved gradually in response to feedback from a community of users (like most successful things). The value of backward compatibility totally overwhelms the minor advantages/disadvantages stemming from this being a reference or a pointer.

Why is 'this' a pointer and not a reference?

2 Answers