Why is 'this' a so-called "rvalue"? Because it is not any pointer or any variable.
Topics
About: C++
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: 'this' Is Not Any Pointer, but a Keyword That Represents an Address
- 2: The Conclusion and Beyond
Starting Context
- The reader has a basic knowledge on C++, even if he or she doesn't accurately understand its some widely-misrepresented elements.
- The reader has a knowledge on what the author distinctly means by 'datum', 'variable', 'expression', and 'value'.
- The reader has a knowledge on what so-called "lvalue" and "rvalue" are in C++.
Target Context
- The reader will understand that 'this' is not any pointer (or any variable), but a keyword that represents an address.
Orientation
Hypothesizer 7
Most tutorials (whether they are popular or not) casually state that 'this' is a pointer, but then, why does this code cause a compile error like "lvalue required as unary ‘&’ operand"?
@C++ Source Code
#include <iostream>class Test1 {void test1 ();};void Test1::test1 () {::std::cout << "The address of 'this' is '" << &this << "'." << ::std::endl << ::std::flush;}
"lvalue required"? . . . Huh? . . . If 'this' was a pointer, the expression, "this", should be an lexpression . . .
Let me clarify a thing: 'pointer' is a variable whose value is an address, and any variable used as an expression by itself is an lexpression.
So, there must be a fallacy that has to be rectified, somewhere.
Main Body
1: 'this' Is Not Any Pointer, but a Keyword That Represents an Address
Hypothesizer 7
In fact, 'this' is not any pointer (or any variable), but a keyword that represents an address.
Huh? What is the difference? . . . Well, if someone does not understand the difference, he or she should not understand what 'variable' is.
For example, an expression, "(Test1 * const) 1", is not any pointer (or any variable), but an expression that represents an address, and certainly, is an rexpression.
Although I do not blindly accept what the C++ standard says (because it includes some incongruous explanations), it gives a reasonable explanation on 'this': "the keyword this is a non-lvalue expression whose value is the address of the object for which the function is called".
2: The Conclusion and Beyond
Hypothesizer 7
Now, I seem to understand that 'this' is not any pointer, but a keyword that represents an address.
As there are some other unsatisfactory explanations on C++, I will try to make more reasonable explanations in future articles.
