Post-Increment and Pre-Increment Overloading

[jehugaleahsa]

Hello:

In C++, you had to distinguish between post and pre increments when
overloading. Could someone give me a short demonstration of how to
write these?

I get the impression that are handled with the same overload, but I
just want to make sure.

[Peter Duniho]

On 2007-11-11 17:10:07 -0800, "je**********@g mail.com"
<je**********@g mail.comsaid:

Hello:

In C++, you had to distinguish between post and pre increments when
overloading. Could someone give me a short demonstration of how to
write these?

How to write which? The C++ versions? Or the C# version?

I get the impression that are handled with the same overload, but I
just want to make sure.

As far as I know, your impression is correct. There is a single ++
operator you can overload. Until you brought up the C++ behavior, I
never even gave much thought to the possibility it might be otherwise.
:)

See http://msdn2.microsoft.com/en-us/library/8edha89s.aspx for more
details on which operators can be overloaded and how to go about it.

Pete

[David Connet]

Peter Duniho <Np*********@Nn OwSlPiAnMk.comw rote in
news:2007111117 592211272-NpOeStPeAdM@NnO wSlPiAnMkcom:

On 2007-11-11 17:10:07 -0800, "je**********@g mail.com"
<je**********@g mail.comsaid:

>Hello:

In C++, you had to distinguish between post and pre increments when
overloading. Could someone give me a short demonstration of how to
write these?

How to write which? The C++ versions? Or the C# version?

>I get the impression that are handled with the same overload, but I
just want to make sure.

As far as I know, your impression is correct. There is a single ++
operator you can overload. Until you brought up the C++ behavior, I
never even gave much thought to the possibility it might be otherwise.
:)

See http://msdn2.microsoft.com/en-us/library/8edha89s.aspx for more
details on which operators can be overloaded and how to go about it.

Pete

I don't know about .Net-land, but in native C++, the pre/post-increment
operators are 2 separate operators.

class xyzzy
{
....
xyzzy& operator++() // prefix
{
// todo: increment this
return *this;
}
xyzzy operator++(int) // postfix
{
xyzzy tmp(*this);
// todo: increment this
return tmp;
}
};

Dave Connet

[Peter Duniho]

On 2007-11-11 19:37:31 -0800, David Connet <co****@entelos .comsaid:

I don't know about .Net-land, but in native C++, the pre/post-increment
operators are 2 separate operators.

Yes, of course they are. Both the OP and I acknowledged that. And I
suppose if the OP is asking for how to declare such in C++, your post
would be relevant to that question (but not what I'd call on-topic in
this newsgroup).

[jehugaleahsa]

On Nov 11, 8:44 pm, Peter Duniho <NpOeStPe...@Nn OwSlPiAnMk.comw rote:

On 2007-11-11 19:37:31 -0800, David Connet <con...@entelos .comsaid:

I don't know about .Net-land, but in native C++, the pre/post-increment
operators are 2 separate operators.

Yes, of course they are. Both the OP and I acknowledged that. And I
suppose if the OP is asking for how to declare such in C++, your post
would be relevant to that question (but not what I'd call on-topic in
this newsgroup).

I was curious about how it is done in C#. I am more than aware of how
to do it in C++.

Thanks for your replies.

[David Connet]

Peter Duniho <Np*********@Nn OwSlPiAnMk.comw rote in
news:2007111119 444827544-NpOeStPeAdM@NnO wSlPiAnMkcom:

On 2007-11-11 19:37:31 -0800, David Connet <co****@entelos .comsaid:

>I don't know about .Net-land, but in native C++, the pre/post-increment
operators are 2 separate operators.

Yes, of course they are. Both the OP and I acknowledged that. And I
suppose if the OP is asking for how to declare such in C++, your post
would be relevant to that question (but not what I'd call on-topic in
this newsgroup).

Sorry - I misread the original post. I thought the OP was asking how to do
it in C++ and C#.

Dave

[Lew]

David Connet wrote:

Sorry - I misread the original post. I thought the OP was asking how to do
it in C++ and C#.

Your interpretation was not contradicted by anything in the original post.

--
Lew

[Chris]

In C++, you had to distinguish between post and pre increments when

overloading. Could someone give me a short demonstration of how to
write these?

I get the impression that are handled with the same overload, but I
just want to make sure.

Since no one has yet answered your question, yes there is only one overload
for both pre and post-increment.

public static EttMonth operator ++(EttMonth m1)
{
return m1 + 1;
}

This is from some of my code. EttMonth is a struct (represents a month in
yyyyMM integer form). The + operator is also overloaded to add the number of
months specified by the RHS.

Chris

[Ben Voigt [C++ MVP]]

<je**********@g mail.comwrote in message
news:11******** *************@5 0g2000hsm.googl egroups.com...

On Nov 11, 8:44 pm, Peter Duniho <NpOeStPe...@Nn OwSlPiAnMk.comw rote:

>On 2007-11-11 19:37:31 -0800, David Connet <con...@entelos .comsaid:

I don't know about .Net-land, but in native C++, the pre/post-increment
operators are 2 separate operators.

Yes, of course they are. Both the OP and I acknowledged that. And I
suppose if the OP is asking for how to declare such in C++, your post
would be relevant to that question (but not what I'd call on-topic in
this newsgroup).

I was curious about how it is done in C#. I am more than aware of how
to do it in C++.

C# doesn't have compound assignment operators (like +=, *=, ++). Instead,
it calls the base operator (+, *) and then the compiler generates the
assignment. Prefix and postfix are handled the same way (see section 7.5.9
of the standard). The operator++ implementation is NOT responsible for
changing the value in-place as it is in C++, rather the C# compiler assigns
the value returned from the operator++ implementation as a separate step.

"The run-time processing of a postfix increment or decrement operation of
the form x++ or x-- consists of the following steps:
· If x is classified as a variable:

o x is evaluated to produce the variable.

o The value of x is saved.

o The selected operator is invoked with the saved value of x as its
argument.

o The value returned by the operator is stored in the location given
by the evaluation of x.

o The saved value of x becomes the result of the operation.

· If x is classified as a property or indexer access:

o The instance expression (if x is not static) and the argument list
(if x is an indexer access) associated with x are evaluated, and the results
are used in the subsequent get and set accessor invocations.

o The get accessor of x is invoked and the returned value is saved.

o The selected operator is invoked with the saved value of x as its
argument.

o The set accessor of x is invoked with the value returned by the
operator as its value argument.

o The saved value of x becomes the result of the operation.

The ++ and -- operators also support prefix notation (§7.6.5). The result of
x++ or x-- is the value of x before the operation, whereas the result of ++x
or --x is the value of x after the operation. In either case, x itself has
the same value after the operation.

An operator ++ or operator -- implementation can be invoked using either
postfix or prefix notation. It is not possible to have separate operator
implementations for the two notations."

>
Thanks for your replies.

[jehugaleahsa]

On Nov 12, 9:55 am, "Ben Voigt [C++ MVP]" <[e-mail] spamwrote:

<jehugalea...@g mail.comwrote in message

news:11******** *************@5 0g2000hsm.googl egroups.com...

On Nov 11, 8:44 pm, Peter Duniho <NpOeStPe...@Nn OwSlPiAnMk.comw rote:

On 2007-11-11 19:37:31 -0800, David Connet <con...@entelos .comsaid:

I don't know about .Net-land, but in native C++, the pre/post-increment
operators are 2 separate operators.

Yes, of course they are. Both the OP and I acknowledged that. And I
suppose if the OP is asking for how to declare such in C++, your post
would be relevant to that question (but not what I'd call on-topic in
this newsgroup).

I was curious about how it is done in C#. I am more than aware of how
to do it in C++.

C# doesn't have compound assignment operators (like +=, *=, ++). Instead,
it calls the base operator (+, *) and then the compiler generates the
assignment. Prefix and postfix are handled the same way (see section 7.5..9
of the standard). The operator++ implementation is NOT responsible for
changing the value in-place as it is in C++, rather the C# compiler assigns
the value returned from the operator++ implementation as a separate step.

"The run-time processing of a postfix increment or decrement operation of
the form x++ or x-- consists of the following steps:
· If x is classified as a variable:

o x is evaluated to produce the variable.

o The value of x is saved.

o The selected operator is invoked with the saved value of x as its
argument.

o The value returned by the operator is stored in the location given
by the evaluation of x.

o The saved value of x becomes the result of the operation.

· If x is classified as a property or indexer access:

o The instance expression (if x is not static) and the argument list
(if x is an indexer access) associated with x are evaluated, and the results
are used in the subsequent get and set accessor invocations.

o The get accessor of x is invoked and the returned value is saved.

o The selected operator is invoked with the saved value of x as its
argument.

o The set accessor of x is invoked with the value returned by the
operator as its value argument.

o The saved value of x becomes the result of the operation.

The ++ and -- operators also support prefix notation (§7.6.5). The result of
x++ or x-- is the value of x before the operation, whereas the result of ++x
or --x is the value of x after the operation. In either case, x itself has
the same value after the operation.

An operator ++ or operator -- implementation can be invoked using either
postfix or prefix notation. It is not possible to have separate operator
implementations for the two notations."

Thanks for your replies.- Hide quoted text -

- Show quoted text -

Well that brings me to my next question - does it bother anyone that
overloaded operators operate on a new instance rather than the given
instance? How do unary operators affect the given instance? If the
operation simply assigns a modified version to my instance, how do I
know the assignment worked as expected? Are these questions worded
well enough?