C / C++ / MFC

George_George wrote:

It is not correct code.

George_George wrote:

Because the destructor of auto_ptr will use delete other than delete[]. And it will lead to memory leak.

auto_ptr<int[10]> pi(new int[10]);

CPallini wrote:

George_George wrote:
Because the destructor of auto_ptr will use delete other than delete[]. And it will lead to memory leak.


FALSE. It allocates just 1 integer (and initialise it to 10) hence no memory leaks after delete.

CPallini wrote:

BTW: no compiler error for the code below

auto_ptr<int[10]> pi(new int[10]);

1>Compiling...
1>main.cpp
1>d:\visual studio 2008\projects\test0401\test0401\main.cpp(7) : error C2664: 'std::auto_ptr<_Ty>::auto_ptr(_Ty (*)) throw()' : cannot convert parameter 1 from 'int *' to 'int (*)[10]'
1>        with
1>        [
1>            _Ty=int [10]
1>        ]
1>        Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast

George_George wrote:

I am confused. We are talking about destructor of auto_ptr? Why do you say "allocates"?? Could you provide more description please?

auto_ptr<int> p(new int(10)); 

George_George wrote:

No, I have compile errors in MSVC 2008.

warning C4156: deletion of an array expression without using the array form of 'delete'; array form substituted
[...].

George_George wrote:

Means auto_ptr will use delete other than delete[] is destructor of auto_ptr?

George_George wrote:

What means "array form substituted"?

_Ty=int [10]

_Ty=int [10]

auto_ptr<int[10]>

George_George wrote:

I think your code will have memory leak potentially? Because delete other than delete[] will work on the array?

George_George wrote:

What is your reply to my original question?

George_George wrote:

You mean your code will cause undefined behavior? CPallini?

auto_ptr<Foo> pi[3]={auto_ptr<Foo>(new Foo), auto_ptr<Foo>(new Foo), auto_ptr<Foo>(new Foo)};