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Why are there throw() following the delete[]?

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  Why are there throw() following the delete[]? fl <rxjwg98@gmail.com> - 2015-07-15 16:05 -0700
    Re: Why are there throw() following the delete[]? Ian Collins <ian-news@hotmail.com> - 2015-07-16 11:18 +1200
      Re: Why are there throw() following the delete[]? fl <rxjwg98@gmail.com> - 2015-07-15 16:36 -0700
        Re: Why are there throw() following the delete[]? Ian Collins <ian-news@hotmail.com> - 2015-07-16 11:44 +1200
        Re: Why are there throw() following the delete[]? Bo Persson <bop@gmb.dk> - 2015-07-16 11:03 +0200

#37090 — Why are there throw() following the delete[]?

Hi,

When I search delete description for C++, I find the following:




ordinary (1)	
void operator delete[] (void* ptr) throw();
nothrow (2)	
void operator delete[] (void* ptr, const std::nothrow_t& nothrow_constant) throw();
placement (3)	
void operator delete[] (void* ptr, void* voidptr2) throw();


The description for the above definitions is:

(1) ordinary delete
Deallocates the memory block pointed to by ptr (if not null), releasing the storage space previously allocated to it by a call to operator new[] and rendering that pointer location invalid.
(2) nothrow delete
Same as above (1).
(3) placement delete
Does nothing.


I am puzzled about:

1. In the normal use of delete[], I never see the throw() following it;
2. The description does not mention about throw() although the upper part
   has throw()


Could you explain it to me as I am still a newbie on such a question? 


Thanks,

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#37091

fl wrote:
> Hi,
>
> When I search delete description for C++, I find the following:
>
> ordinary (1)	
> void operator delete[] (void* ptr) throw();
> nothrow (2)	
> void operator delete[] (void* ptr, const std::nothrow_t& nothrow_constant) throw();
> placement (3)	
> void operator delete[] (void* ptr, void* voidptr2) throw();
>
<snip>
>
> I am puzzled about:
>
> 1. In the normal use of delete[], I never see the throw() following it;
> 2. The description does not mention about throw() although the upper part
>     has throw()

These describe the operators as defined in <new>.  The throw() (noexcept 
in modern C++) is telling you that the operators will not throw.

You don't include the throw() when you use operator delete, it is only 
appropriate in the declaration.


-- 
Ian Collins

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#37094

On Wednesday, July 15, 2015 at 4:18:40 PM UTC-7, Ian Collins wrote:
> fl wrote:
> > Hi,
> >
> > When I search delete description for C++, I find the following:
> >
> > ordinary (1)	
> > void operator delete[] (void* ptr) throw();
> > nothrow (2)	
> > void operator delete[] (void* ptr, const std::nothrow_t& nothrow_constant) throw();
> > placement (3)	
> > void operator delete[] (void* ptr, void* voidptr2) throw();
> >
> <snip>
> >
> > I am puzzled about:
> >
> > 1. In the normal use of delete[], I never see the throw() following it;
> > 2. The description does not mention about throw() although the upper part
> >     has throw()
> 
> These describe the operators as defined in <new>.  The throw() (noexcept 
> in modern C++) is telling you that the operators will not throw.
> 
> You don't include the throw() when you use operator delete, it is only 
> appropriate in the declaration.
> 
> 
> -- 
> Ian Collins

Thanks. As you said, it is a declaration. Good.
As I am a newbie to C++, I still feel it is different from a normal 
declaration. What role of the throw() in the declaration?
I hope your further explanation can teach me more on this.
Thanks again.

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#37095

fl wrote:
> On Wednesday, July 15, 2015 at 4:18:40 PM UTC-7, Ian Collins wrote:
>> fl wrote:
>>> Hi,
>>>
>>> When I search delete description for C++, I find the following:
>>>
>>> ordinary (1)	
>>> void operator delete[] (void* ptr) throw();
>>> nothrow (2)	
>>> void operator delete[] (void* ptr, const std::nothrow_t& nothrow_constant) throw();
>>> placement (3)	
>>> void operator delete[] (void* ptr, void* voidptr2) throw();
>>>
>> <snip>
>>>
>>> I am puzzled about:
>>>
>>> 1. In the normal use of delete[], I never see the throw() following it;
>>> 2. The description does not mention about throw() although the upper part
>>>      has throw()
>>
>> These describe the operators as defined in <new>.  The throw() (noexcept
>> in modern C++) is telling you that the operators will not throw.
>>
>> You don't include the throw() when you use operator delete, it is only
>> appropriate in the declaration.
>
> Thanks. As you said, it is a declaration. Good.
> As I am a newbie to C++, I still feel it is different from a normal
> declaration. What role of the throw() in the declaration?

As I said above, throw() (noexcept in modern C++) is telling you that 
the operators will not throw an exception.

-- 
Ian Collins

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#37102

On 2015-07-16 01:36, fl wrote:
> On Wednesday, July 15, 2015 at 4:18:40 PM UTC-7, Ian Collins wrote:
>> fl wrote:
>>> Hi,
>>>
>>> When I search delete description for C++, I find the following:
>>>
>>> ordinary (1)	
>>> void operator delete[] (void* ptr) throw();
>>> nothrow (2)	
>>> void operator delete[] (void* ptr, const std::nothrow_t& nothrow_constant) throw();
>>> placement (3)	
>>> void operator delete[] (void* ptr, void* voidptr2) throw();
>>>
>> <snip>
>>>
>>> I am puzzled about:
>>>
>>> 1. In the normal use of delete[], I never see the throw() following it;
>>> 2. The description does not mention about throw() although the upper part
>>>      has throw()
>>
>> These describe the operators as defined in <new>.  The throw() (noexcept
>> in modern C++) is telling you that the operators will not throw.
>>
>> You don't include the throw() when you use operator delete, it is only
>> appropriate in the declaration.
>>
>>

> Thanks. As you said, it is a declaration. Good.
> As I am a newbie to C++, I still feel it is different from a normal
> declaration. What role of the throw() in the declaration?
> I hope your further explanation can teach me more on this.
>


The throw-part lists the execeptions the function (or operator) is 
allowed to throw. Here the list is empty, meaning that it throws no 
exceptions.



Bo Persson

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