Re: Order of destructor execution.

From:
 James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 24 Jul 2007 12:37:01 -0000
Message-ID:
<1185280621.733264.45960@q75g2000hsh.googlegroups.com>
On Jul 23, 7:24 pm, "Victor Bazarov" <v.Abaza...@comAcast.net> wrote:

eshn...@gmail.com wrote:

[..]
unsigned getCont( void )
{
  MutexLocker( mut );


Not sure what 'mut' here is,


It doesn't matter.

but if it's a global object of some
kind, then the statement above creates (and immediately discards)
a temporary value of type 'MutexLocker'.


No. That statement declares a variable mut, of type
MutexLocker. If MutexLocker doesn't have a default constructor,
it is illegal, and the compiler should complain. If MutexLocker
does have a default constructor, then it is called.

The destructor for that
temporary is called before the following 'return' statement is
exectuted.


The destructor of mut is called on leaving the function.

  return( cont );
}

Will the destructor ~MutexLocker() be executed before the copy
constructor of unsigned?


There is no "copy constructor of unsigned". Does all that clarify
the situation?


Conceptually, cont can be copied, as if it had a copy
constructor. The destructor of the variable mut will be called
after this copy.

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