Re: Memory leak
From: Karl Heinz Buchegger (kbuchegg_at_gascad.at)
Date: 05/17/04
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Date: Mon, 17 May 2004 11:19:11 +0200
John wrote:
>
> Does it mean the number of "new" is equal to the number of "delete" in
> a code?
> I think there might be exceptions, for example,
>
> void f(X* x0)
> {
> X* x1 = new X;
> X* x2 = new X;
> X* x3 = new X;
> ...
> x1 = x0;
> x2 = x0;
> x3 = x0;
> }
>
> main()
> {
> ....
> f(x0);
> ....
> f1(x0);
>
> }
>
> In the above code, x0 exists outside f(). If I "delete" anyone of x1,
> x2 and x3 before the end of f(), x0 will be released. So f1() will not
> run correctly.
> So there is less "delete" than "new".
> Am I right?
When you run this code, the following is happening:
in main there exists x0
x0
+-----------+
| |
+-----------+
then comes the function call to f.
In f new variables are created and initalized
X* x1 = new X;
X* x2 = new X;
X* x3 = new X;
That means:
x0
+-----------+
| |
+-----------+
f::x1
+-------+ +-------------+
| o------------------------------->| |
+-------+ +-------------+
f::x2
+-------+ +-------------+
| o------------------------------->| |
+-------+ +-------------+
f::x3
+-------+ +-------------+
| o------------------------------->| |
+-------+ +-------------+
And now you do:
x1 = x0; // x0 beeing the address of variable x0 in main
x0
+-----------+
| |<+
+-----------+ |
|
|
f::x1 |
+-------+ | +-------------+
| o---------+ | |
+-------+ +-------------+
You are certainly right, that you must not do a delete on x1, since this
will try to free the memory occupied by variable x0. But look at the
lonely rectangle. There is no pointer pointing to it anymore, which
means you have no way of freeing it again: you created a memory leak.
The sequence:
delete x1;
x1 = x0;
on the other hand will do:
This was the starting situation after the allocations
x0
+-----------+
| |
+-----------+
f::x1
+-------+ +-------------+
| o------------------------------->| |
+-------+ +-------------+
delete x1
x0
+-----------+
| |
+-----------+
f::x1
+-------+
| o------------------------------->
+-------+
x1 = x0;
x0
+-----------+
| |<+
+-----------+ |
|
|
f::x1 |
+-------+ |
| o---------+
+-------+
and as you can see, there is no longer a rectangle sitting
in memory with no pointer to it: no memory leak.
So the correct sequence in your example would be:
void f(X* x0)
{
X* x1 = new X;
X* x2 = new X;
X* x3 = new X;
...
delete x1;
x1 = x0;
delete x2;
x2 = x0;
delete x3;
x3 = x0;
}
And as you can see, at runtime there is exactly one delete
for every new.
-- Karl Heinz Buchegger kbuchegg@gascad.at
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