Format of compiler generated derived destructor when base has 'virtual ~base() throw():"



 DEVELOP > c-Plus-Plus > Format of compiler generated derived destructor when base has 'virtual ~base() throw():"

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Topic: DEVELOP > c-Plus-Plus
User: "qazmlp"
Date: 08 Apr 2005 05:26:06 AM
Object: Format of compiler generated derived destructor when base has 'virtual ~base() throw():"
I have a class like this:
class base
{
public:
virtual ~base() throw();
// Other members
};
Now, if I write a class like this:
class derived:public base
{
// Other members
};
What will/should be the format of the compiler-generated destructor for derived?
Will it be this:
- ~derived(); or
- virtual ~derived(); or
- virtual ~derived() throw();
.

User: "qazmlp"

Title: Re: Format of compiler generated derived destructor when base has 'virtual ~base() throw():" 10 Apr 2005 10:09:07 AM
Could anybody reply to this post? I would like to know what exactly
'C++ standard' mentions about this.
Thanks!

qazmlp1209@rediffmail.com (qazmlp) wrote in message
news:<db9bbf31.0504080226.71c45717@posting.google.com>...
I have a class like this:
class base
{
public:
virtual ~base() throw();
// Other members
};


Now, if I write a class like this:
class derived:public base
{
// Other members
};

What will/should be the format of the compiler-generated destructor for derived?

Will it be this:
- ~derived(); or
- virtual ~derived(); or
- virtual ~derived() throw();

.


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