Can we override constructor

[ad]

When a class (say class1) inherited form another class (Class0).

The constructor in the class0 will be executing before the constructor of
class1.

Can we over the constructor?

[Amit]

Depends on what kind of constructor it is - if it is the default constructor
(i.e. constructor with no parameters) then you can't override the parent's
constructor, it will always execute. If it is a non-default constructor then
it is automatically overriden unless you call it using the "base" keyword.
Note that the default constructor of the parent is always called regardless
of what kind of constructor is being invoked of the child class. Also,
whenever a base-class constructor executes, it does so before the inherited
class constructor is run.
Hope that helps,
~amit

"ad" <[e-mail]> wrote in message
news:uJ**************@TK2MSFTNGP09.phx.gbl...

When a class (say class1) inherited form another class (Class0).

The constructor in the class0 will be executing before the constructor of
class1.

Can we over the constructor?

[Jon Shemitz]

Amit wrote:

Note that the default constructor of the parent is always called regardless
of what kind of constructor is being invoked of the child class.

No, this is not true. You can specify which base constructor will be
called -

public ThisType(t1 p1, t2 p2): base(p1) { /*...*/ }

--

programmer, author http://www.midnightbeach.com
and father http://www.midnightbeach.com/hs

[Amit]

oops you're right... the default constructor of the parent class is always
called IF no other constructor is specified.
"Jon Shemitz" <jo*@midnightbeach.com> wrote in message
news:41***************@midnightbeach.com...

Amit wrote:

Note that the default constructor of the parent is always called regardless of what kind of constructor is being invoked of the child class.

No, this is not true. You can specify which base constructor will be
called -

public ThisType(t1 p1, t2 p2): base(p1) { /*...*/ }

--

programmer, author http://www.midnightbeach.com
and father http://www.midnightbeach.com/hs

[Jon Skeet [C# MVP]]

ad <[e-mail]> wrote:

When a class (say class1) inherited form another class (Class0).

The constructor in the class0 will be executing before the constructor of
class1.

Can we over the constructor?

You can't override constructors, because they aren't inherited. It's
not entirely clear what you're after, but the following page may help:

http://www.pobox.com/~skeet/csharp/constructors.html

--
Jon Skeet - <sk***@pobox.com>
http://www.pobox.com/~skeet
If replying to the group, please do not mail me too

[Jon Shemitz]

Amit wrote:

oops you're right... the default constructor of the parent class is always
called IF no other constructor is specified.

Yes, this is better. A base constructor will always be called, for
every class that descends from object, because every class must have
at least one constructor that calls a base() constructor (explicitly
or implicitly) and every call to a this() constructor must ultimately
call a base() constructor.

--

programmer, author http://www.midnightbeach.com
and father http://www.midnightbeach.com/hs

[James Curran]

"Jon Shemitz" <jo*@midnightbeach.com> wrote in message
news:41***************@midnightbeach.com...

A base constructor will always be called, for
every class that descends from object, because every class must have
at least one constructor that calls a base() constructor (explicitly
or implicitly)

And because, one way or another, the base object must be constructed.

--
Truth,
James Curran
[erstwhile VC++ MVP]
Home: www.noveltheory.com Work: www.njtheater.com
Blog: www.honestillusion.com Day Job: www.partsearch.com

[Jon Shemitz]

James Curran wrote:

"Jon Shemitz" <jo*@midnightbeach.com> wrote in message
news:41***************@midnightbeach.com...

A base constructor will always be called, for
every class that descends from object, because every class must have
at least one constructor that calls a base() constructor (explicitly
or implicitly)

And because, one way or another, the base object must be constructed.

Yes. That's the semantic reason for the syntactic requirement.

--

programmer, author http://www.midnightbeach.com
and father http://www.midnightbeach.com/hs