Re: Order of destructor execution.

"Jim Langston" <>
Mon, 23 Jul 2007 17:55:34 -0700
<> wrote in message

Sorry if this may seem to be very obvious, but it is important and I
decided to ask.

I have the following classes:

class MutexWrapper
     MutexWrapper( void );
     void lock( void ); //Maybe const, whatever.
     void unlock( void ); //As well.
     ~MutexWrapper( void );

class MutexLocker
     MutexLocker( MutexWrapper& mut ){ mut.lock(); } //Perhaps it
could be const MutexWrapper& mut (?)
     ~MutexLocker( void ){ mut.unlock(); }

So that I want to sync acces to the variable
volatile unsigned cont
in the following piece of code (supposing there is a global
MutexWrapper mut ):

unsigned getCont( void )
  MutexLocker( mut );

You are not storing the instance of MutexLocker anywhere. It is created,
then prompty destroyed.

  return( cont );

At this point your MutexLocker is already destroyed.


Will the destructor ~MutexLocker() be executed before the copy
constructor of unsigned? It seems logical if this return statement
could be interpreted as a placement new in the adress given by the
return variable, something like the equivalence of:

c = getCont();


  MutexLocker( mut );
  new( &c ) unsigned( cont );

If so, my code seems to be correct. Is it true?

You need to store your MutexLocker somewhere.

unsigned getCont( void )
    MutexLocker LockIt( mut );
    return( cont );

By giving the MutexLocker a variable name, its lifetime is the lifetime of
the function/method.

The quesiton still becomes, which is done first, the destruction of LockIt
or the retriving of the cont value? I'm not really sure, someone might
know. But being unsure I would do this.

unsigned getCont( void )
    MutexLocker LockIt( mut );
    unsigned ReturnVal = cont;
    return( ReturnVal );

I know for a fact that ReturnVal is going to be assigned while the instance
of the MutexLocker is still there. It may not be necessary, but I know it
would work.

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