重载前增量和后增量

Question

I saw an example about implementing pre-increment and post-increment, which claims that overloading pre-increment is able to be defined as

T& T ::operator++()

and overloading post-increment can be defined and implemented in terms of pre-incremet as follows

const T T::operator++(int){
  const T old(*this);
  ++(*this);
  return old;
}

I have two questions:

1) what does “old” mean?

2) ++(*this) is assumed to use the pre-increment, and the original pre-increment definition does not have argument. However, it has *this here.

Answer 1

what does "old" mean?

The method is a post increment. The current value ("old value") is returned and then the value is incremented ("new value").

++(*this) is assumed to use the pre-increment, and the original pre-increment definition does not have argument. However, it has *this here.

*this is not an argument. The parentheses are not necessary, they are there for readability.
It's equivalent to ++*this .

Answer 2

1) "old" is the value that "this" had before it was incremented. Post-increment is supposed to return that value.

2) ++(*this) is equivalent to this->operator++()

Answer 3

2) ++(*this) is assumed to use the pre-increment, and the original pre-increment definition does not have argument. However, it has *this here.

++ is a unary operator, so it has an argument alright. Whenever you overload an operator as a member function, the first argument is the current object.

The unused int parameter is but a hack to distinguish between pre- and post-increment, because operator++ can mean either. Post-increment does not really accept an integer*, it is just a (awkward) language construct.

You can also overload these operators as free functions:

struct T
{
    int n;
};

T& operator++(T& t) { ++t.n; return t; }
T operator++(T& t, int) { T old(t); ++t; return old; }

int main()
{
   T a;
   T b = a++;
   ++b;
}

*under normal usage. When you invoke the operator using function call syntax, you can pass an additional int to distinguish between those two:

operator++(a); //calls-preincrement
operator++(b, 1); //calls post-increment

Answer 4

old is simply a variable name (it's not a keyword, if that's what you were wondering). It's used to save the previous value of the operand.