Re: returning dynamically allocated object through reference

From:
Thomas Maeder <maeder@glue.ch>
Newsgroups:
comp.lang.c++.moderated
Date:
Fri, 23 Jan 2009 11:41:39 CST
Message-ID:
<m21vuve0gz.fsf@glue.ch>
mundacho@gmail.com writes:

I'd like to know if this code is going to create a memory leak in
the calling function (this codes compiles):


It will create a memory leak.

ili::IntImg
&RectangleImageComparator::synchronize(const ili::IntImg &pImage)
   {
     ili::IntImg *tempObj = new ili::IntImg(pImage);
     return *tempObj;
   }

i wanted to return a reference because it's faster because the
object is not copied but passed directly, isn't it?.


It's faster if you measure and your measurements say it's faster.

Unless supported by measuremenets, such attempts at optimization very
often target code that isn't relevant for the performance of the
program.

My question is if the calling function will free automatically the
object returned, if I do:

// in the calling function I'd do:
ili::IntImg myVar = synchronize(image); // will myVar be destroyed
automatically???


Didn't you write that you wanted to save the copying? Then why do you
copy now?

And no, the object created in synchronize() is not destructed. Doing
so would be disastrous if synchronize returned a reference to an
object that was not dynamically created, which the caller can't tell.

Consider
- returning from synchronize() by value
- returning a smart pointer object

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