Mail Archives: djgpp/1997/06/17/02:46:06
Andrew Deren <aderen AT eecs DOT uic DOT edu> wrote in article
<Pine DOT SUN DOT 3 DOT 95 DOT 970613073825 DOT 25529B-100000 AT ernie DOT eecs DOT uic DOT edu>...
[extraneous stuff snipped throughout]
> > my question is:
> > will the constructor generate a new copy of this object each time I
call
> > it.
> > And if it does will the memory previously taken by my_obj be freed?
> my_obj Object;
This needs to be changed to "Object my_obj;"
This will result in the allocation of storage for an instance of the class
"Object"
named "my_obj". It will also cause the invocation of the default
constructor
(the one that takes no parameters) for objects of the Object class. This
constructor will get called before the "main()" function is entered.
> void InitObj(int x, int y)
> {
> my_obj = Object(x, y);
This line will cause the invocation of the constructor for the "Object"
class that takes two intergers as parameters; it will construct an
anonymous instance of the "Object" class. This anonymous instance
is then the parameter for the assignment operator that is called to
assign its value to the my_obj instance of the Object class.
The memory previously taken by my_obj won't be
freed when this assignment operator executes.
And sooner or later you'll come to grief on that very fact. Probably
sooner, because you'll add a "pointer-to-thingy" member to your
"Object" class and call "new" or "malloc" to allocate some memory
and then wonder why every call to InitObj() seems to loose some
memory. The answer is: default assignment operators do memberwise
assignment, which means that your "pointer-to-thingy" member gets
assigned the value WITHOUT THE PREVIOUS value getting "delete"
or "free()" called for it. If you want to know more about this, get a
good C++ introductory book.
> }
This (the end-of-function brace for the InitObj() function) will cause
the destructor for the "Object" class to get called for the anonymous
instance of "Object" created above.
--
John Bodfish
bodfish AT als DOT ameritech DOT com
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