[Cs3] operator++

[Mikhail Nesterenko]

> 
> In class we talked about how something like
>
> Int i = 0;
> 
> ++i++;
> }
> Is syntactically illegal in C++, however I was wondering how C++ handles
> {
> Int i = 0;
> i.operator++(i++);
> }
> 

operator++ does not accept integer as a parameter. Well, it does, but
it is a "dummy" integer (to be ignored) to distinguish the signatures
of prefix and postifx version of increment.

That is, the signatures are:

     Number& operator++ ();    // prefix ++
     Number  operator++ (int); // postfix ++

See this article in C++ FAQ for more info
   http://www.parashift.com/c++-faq-lite/increment-pre-post-overloading.html

So, the above would not do what you are thinking anyway. Moreover, it
would fail to compile because "i" is declared integer (assuming "int"
is lower case) which is a primitive type for which dot-operator is not
applicable.

However, what I wanted to clarify is the difference between prefix and
postfix forms of the increment. In C++, prefix form returns an l-value
(assignable) while postfix form returns an r-value (not
assignable). However, both prefix and postfix forms need l-value as an
argument.  Therefore, prefix form is stackable while postfix form is
not:

  ++++++i; // legal
  i++++++; // not legal

The case of 

  ++i++;

is curious. The precedence of the postfix form is higher than that of
the prefix form. Hence, in the above statement, postfix is evaluated
first, it returns an r-value, which is not allowed as an argument for
prefix form of the increment.

However,

 (++i)++; // is legal

Below is an example program that demonstrates using the prefix form of
the increment as an l-value.

Thanks,
--
Mikhail

----------------------------------
// demonstrates that pre-increment returns an l-value
// Mikhail Nesterenko
// 1/15/2014

#include <iostream>

int main(void){

  int i=0;

  ++i = 22 + 5;
  (++i)++;

  std::cout << ++--++i << std::endl;
}