Memory leak

[John]

Hi all:

When I run my code, I find that the memory that the code uses keeps
increasing.
I have a PC with 2G RAM running Debian linux. The code consumes 1.5G
memory by the time it finishes execution. But I do not think it needs
so much memory. About 500M memory should be enough. I have following
questions about memory leak.
(1).If in my code I only define constructor for my class, and do not
define destructor, will it cause memory leak?
(2).If in my code I only use "new" to declare new object, and do not
use "delete", will it cause memory leak?

For example, in the following code:

void class1::function1()
{
class2 *r1;
class2 *r2 = new class2;
class2 *rr[20];

......

function2(rr);

......
//r1 and r2 are also used in function1().

}

In the above code, I have two classes and I define constructor for the
two classes and do not define destructor. In function1(), I declare
two pointers of class2 and an array of pointer of class2. The array rr
is used to bring back values from function2(). For r2, I do not use
"delete".
Will r1, r2 and the array rr[] cause memory leak?
The two pointers, r1 and r2, and array rr[] are local variables, when
the code exits function1(), these local variables should be released
automatically.
Am I right?

Thanks a lot.

John

 

[John Harrison]

Hi all:

When I run my code, I find that the memory that the code uses keeps
increasing.
I have a PC with 2G RAM running Debian linux. The code consumes 1.5G
memory by the time it finishes execution. But I do not think it needs
so much memory. About 500M memory should be enough. I have following
questions about memory leak.
(1).If in my code I only define constructor for my class, and do not
define destructor, will it cause memory leak?

Depends on the class.

(2).If in my code I only use "new" to declare new object, and do not
use "delete", will it cause memory leak?

Yes.

Its a very simple rule, nothing to do with constructors or destructors. When
your program runs every new allocates some memory, if you don't do a delete
for the same memory then you have a memory leak.

For example, in the following code:

void class1::function1()
{
class2 *r1;
class2 *r2 = new class2;
class2 *rr[20];

......

function2(rr);

......
//r1 and r2 are also used in function1().

}

In the above code, I have two classes and I define constructor for the
two classes and do not define destructor.

That's irrelevant.

In function1(), I declare
two pointers of class2 and an array of pointer of class2. The array rr
is used to bring back values from function2(). For r2, I do not use
"delete".
Will r1, r2 and the array rr[] cause memory leak?

Where are the deletes? There are no deletes so there are memory leaks all
over the place.

The two pointers, r1 and r2, and array rr[] are local variables, when
the code exits function1(), these local variables should be released
automatically.
Am I right?

Wrong. The variables destructed and the memory they occupy is 'released',
BUT the memory they might be pointing to is not released.

void f()
{
X x;
X* xp = new X();
...
}

When you get to the end of this function, the memory for x and xp are both
released. That has nothing to do with the memory pointed to by xp, which is
a completely different thing. The memory for a pointer and the memory that
it points to are not the same thing.

It's very simple, every new must be matched by a delete.

john

 

[Kutty Banerjee]

Hi all:

When I run my code, I find that the memory that the code uses keeps
increasing.
I have a PC with 2G RAM running Debian linux. The code consumes 1.5G
memory by the time it finishes execution. But I do not think it needs
so much memory. About 500M memory should be enough. I have following
questions about memory leak.
(1).If in my code I only define constructor for my class, and do not
define destructor, will it cause memory leak?
(2).If in my code I only use "new" to declare new object, and do not
use "delete", will it cause memory leak?

For example, in the following code:

void class1::function1()
{
class2 *r1;
class2 *r2 = new class2;
class2 *rr[20];

......

function2(rr);

......
//r1 and r2 are also used in function1().

}

In the above code, I have two classes and I define constructor for the
two classes and do not define destructor. In function1(), I declare
two pointers of class2 and an array of pointer of class2. The array rr
is used to bring back values from function2(). For r2, I do not use
"delete".
Will r1, r2 and the array rr[] cause memory leak?
The two pointers, r1 and r2, and array rr[] are local variables, when
the code exits function1(), these local variables should be released
automatically.
Am I right?

Thanks a lot.

John

Hi,
check out "auto_ptr" from any standard c++ boooks.

kutty

 

[Ian]

Hi all:

When I run my code, I find that the memory that the code uses keeps
increasing.
I have a PC with 2G RAM running Debian linux. The code consumes 1.5G
memory by the time it finishes execution. But I do not think it needs
so much memory. About 500M memory should be enough. I have following
questions about memory leak.
(1).If in my code I only define constructor for my class, and do not
define destructor, will it cause memory leak?

See next answer!

(2).If in my code I only use "new" to declare new object, and do not
use "delete", will it cause memory leak?

Yes, without a destructor, how is the memory you allocated in the
destructor freed?

An alternative is to use std::auto_ptr rather than a plain pointer.

As a rule, any class with pointers must have a destructor and copy
constructor (if nothing else, this makes you think about ownership of
the data you reference through a pointer).

Ian

For example, in the following code:

void class1::function1()
{
class2 *r1;
class2 *r2 = new class2;
class2 *rr[20];

......

function2(rr);

......
//r1 and r2 are also used in function1().

}

In the above code, I have two classes and I define constructor for the
two classes and do not define destructor. In function1(), I declare
two pointers of class2 and an array of pointer of class2. The array rr
is used to bring back values from function2(). For r2, I do not use
"delete".
Will r1, r2 and the array rr[] cause memory leak?
The two pointers, r1 and r2, and array rr[] are local variables, when
the code exits function1(), these local variables should be released
automatically.
Am I right?

Thanks a lot.

John

 

[John Harrison]

See next answer!
Yes, without a destructor, how is the memory you allocated in the
destructor freed?

An alternative is to use std::auto_ptr rather than a plain pointer.

As a rule, any class with pointers must have a destructor and copy
constructor (if nothing else, this makes you think about ownership of
the data you reference through a pointer).

Ian

I think the OP is not talking about a class with pointers but a pointer to a
class. Certainly that is what his code shows.

In any case the rule is every new must be matched with a delete.
Constructors and destructors are just a useful way of making sure that this
rule is followed.

Since the OP is clearly struggling with pointers I would advise avoid using
new where possible. It's certainly a common newbie trait to use new where it
isn't necessary.

// newbie style coding
void func()
{
X* xp = new X;
x->some_func();
delete x;
}

// simpler, safer and better coding
void func()
{
X x;
x.some_func();
}

There's nothing in the code that John posted that indicated he must use new
at all.

john

 

[Rolf Magnus]

Hi all:

When I run my code, I find that the memory that the code uses keeps
increasing.
I have a PC with 2G RAM running Debian linux. The code consumes 1.5G
memory by the time it finishes execution. But I do not think it needs
so much memory. About 500M memory should be enough. I have following
questions about memory leak.
(1).If in my code I only define constructor for my class, and do not
define destructor, will it cause memory leak?

That depends on what the class does. If it allocates any resources, it
should have a destructor that deallocates them.

(2).If in my code I only use "new" to declare new object, and do not
use "delete", will it cause memory leak?

Yes. Everything you got from new should be deleted as soon as you don't
need it anymore.

For example, in the following code:

void class1::function1()
{
class2 *r1;
class2 *r2 = new class2;
class2 *rr[20];

......

function2(rr);

......
//r1 and r2 are also used in function1().

}

In the above code, I have two classes and I define constructor for the
two classes and do not define destructor. In function1(), I declare
two pointers of class2 and an array of pointer of class2. The array rr
is used to bring back values from function2(). For r2, I do not use
"delete".
Will r1, r2 and the array rr[] cause memory leak?

If you don't delete them, yes. However, since the pointers are local
variables within your function and not member variables of your class,
you don't need a destructor for cleaning them up, but rather would
delete them in the function. Do you actually need them to be pointers
anyway? If a direct instance suffices, use it.

The two pointers, r1 and r2, and array rr[] are local variables, when
the code exits function1(), these local variables should be released
automatically.

The local variables are, but not the objects they point to.

Am I right?

What you wrote is right, but I don't think it's what you meant

 

[Anil Mamede]

Hi all:

When I run my code, I find that the memory that the code uses keeps
increasing.
I have a PC with 2G RAM running Debian linux. The code consumes 1.5G
memory by the time it finishes execution. But I do not think it needs
so much memory. About 500M memory should be enough. I have following

Maybe you're creating objects inside a infinit while or for loop.

Anil Mamede

 

[JKop]

Where you use "new", you must use "delete".

Where you use "new", you must use "delete".

Where you use "new", you must use "delete".


class2* r2 = new class2;

delete r2;



I highly suggest that you keep the asterisk beside THE CLASS NAME rather
than beside the variable name.

Think of "delete" as a function that takes one paramater, a pointer. You
pass it the value of the variable "r2", which is of type "class2*".

delete r2;


---


As for normal run-of-the-mill variables, they come into being at the
beginning of the block of code and are destroyed at the end of the block of
code:


int main(void)
{ //Right here, a comes into being

class2 a;

{ //Right here, b comes into being

class2 b;

} // Right here, b is destroyed.


return 0;

} //Right here, a is destroyed

 

[Rolf Magnus]

Think of "delete" as a function that takes one paramater, a pointer.
You pass it the value of the variable "r2", which is of type
"class2*".

delete r2;

I'd rather say think of new as a source and delete as a sink. Everything
that came from 'new' has to go to 'delete'.

As for normal run-of-the-mill variables, they come into being at the
beginning of the block

No, they start existing where they are defined.

of code and are destroyed at the end of the
block of code:


int main(void)
{ //Right here, a comes into being

Nope. a doesn't exist yet.

class2 a;

Here, a commes into existance.

{ //Right here, b comes into being

No. Same as a

 

[John]

I think the OP is not talking about a class with pointers but a pointer to a
class. Certainly that is what his code shows.

In any case the rule is every new must be matched with a delete.
Constructors and destructors are just a useful way of making sure that this
rule is followed.

Since the OP is clearly struggling with pointers I would advise avoid using
new where possible. It's certainly a common newbie trait to use new where it
isn't necessary.

// newbie style coding
void func()
{
X* xp = new X;
x->some_func();
delete x;
}

// simpler, safer and better coding
void func()
{
X x;
x.some_func();
}

There's nothing in the code that John posted that indicated he must use new
at all.

john

Hi John,

Thanks a lot.
My code is based on an old code. I modify it and add my own functions.
The old code uses linked list of objects everywhere. The linked list
is implemented by using pointer. This is the reason why I must use
pointer to a class.
If I do not use "new", will it cause memory leak?
For example,

void func()
{
int *p = 0; // Initialize p to be 0.
int a = 10;
X *x; //line 1
x->some_func();
p = & a;
}

Will the above code cause memory leak?
When line 1 is executed, the constructor of X will initialize x. By
the end of func(), x is released. Am I right?

Thanks again.

John

 

[John Harrison]

"John Harrison" <[email protected]> wrote in message

Hi John,

Thanks a lot.
My code is based on an old code. I modify it and add my own functions.
The old code uses linked list of objects everywhere. The linked list
is implemented by using pointer. This is the reason why I must use
pointer to a class.
If I do not use "new", will it cause memory leak?
For example,

void func()
{
int *p = 0; // Initialize p to be 0.
int a = 10;
X *x; //line 1
x->some_func();
p = & a;
}

Will the above code cause memory leak?

No, the above code will crash, x is an uninitalised pointer.

When line 1 is executed, the constructor of X will initialize x.

No, pointers do not have constructors, x is a pointer.

By
the end of func(), x is released. Am I right?

You seem hung up on pointers, often you don't have to use them. Here is the
above program rewritten so that it doesn't use a pointer.

void func()
{
int *p = 0; // Initialize p to be 0.
int a = 10;
X x; // not a pointer
x.some_func(); // no need for a pointer
p = & a;
}

Now because x is not a pointer, there are no problems.

Do not use pointers unless you are sure that you really need them. Linked
lists are one example of where you do need to use pointers, but as you are
finding out using pointers is tricky. If you can I would switch to C++'s
built in linked list class. That should cure your memory leak problems.

john

 

[Rolf Magnus]

My code is based on an old code. I modify it and add my own functions.
The old code uses linked list of objects everywhere. The linked list
is implemented by using pointer. This is the reason why I must use
pointer to a class.
If I do not use "new", will it cause memory leak?
No.

For example,

void func()
{
int *p = 0; // Initialize p to be 0.
int a = 10;
X *x; //line 1
x->some_func();
p = & a;
}

Will the above code cause memory leak?

You can't call some_func() on x, because it points into the desert. You
must first let x point to an object before you can use it.

When line 1 is executed, the constructor of X will initialize x.

No, it won't. You seem to be very confused about objects vs. pointers,
construction vs. memory allocation and about the lifetime of object
created in several different ways. I suggest getting a good book about
C++ that explains those concepts in depth. You won't do yourself any
favor if you stick to this trial/error method.

By the end of func(), x is released. Am I right?

Again, x is a pointer. The memory that x (i.e. the pointer) uses will be
freed, nothing more.
Further, the lifetime of an object is independant of whether a pointer
points to it or not, so in your above example, the memory that a takes
will be released when the function returns, but that doesn't have
anything to do with the existance of p.

 

[JKop]

You are right in that any and all constructors are not called yet, but the
actual memory *is* allocated for the variables at the beginning of the code
block... I think.

 

[Karthik]

Hi all:

When I run my code, I find that the memory that the code uses keeps
increasing.
I have a PC with 2G RAM running Debian linux. The code consumes 1.5G
memory by the time it finishes execution. But I do not think it needs
so much memory. About 500M memory should be enough. I have following
questions about memory leak.
(1).If in my code I only define constructor for my class, and do not
define destructor, will it cause memory leak?
(2).If in my code I only use "new" to declare new object, and do not
use "delete", will it cause memory leak?

For example, in the following code:

void class1::function1()
{
class2 *r1;
class2 *r2 = new class2;
class2 *rr[20];

......

function2(rr);

......
//r1 and r2 are also used in function1().

}

As a thumb rule, make sure that the new and delete operators occur
together in a function body. That makes it easy to maintain and spot the
bugs.

That said, where are you free-ing ( deleting ) the memory allocated
for r2 ( of type class2 ). May be, that might be the culprit.

 

[John]

Hi all:

When I run my code, I find that the memory that the code uses keeps
increasing.
I have a PC with 2G RAM running Debian linux. The code consumes 1.5G
memory by the time it finishes execution. But I do not think it needs
so much memory. About 500M memory should be enough. I have following
questions about memory leak.
(1).If in my code I only define constructor for my class, and do not
define destructor, will it cause memory leak?

Depends on the class.

(2).If in my code I only use "new" to declare new object, and do not
use "delete", will it cause memory leak?

Yes.

Its a very simple rule, nothing to do with constructors or destructors. When
your program runs every new allocates some memory, if you don't do a delete
for the same memory then you have a memory leak.

For example, in the following code:

void class1::function1()
{
class2 *r1;
class2 *r2 = new class2;
class2 *rr[20];

......

function2(rr);

......
//r1 and r2 are also used in function1().

}

In the above code, I have two classes and I define constructor for the
two classes and do not define destructor.

That's irrelevant.

In function1(), I declare
two pointers of class2 and an array of pointer of class2. The array rr
is used to bring back values from function2(). For r2, I do not use
"delete".
Will r1, r2 and the array rr[] cause memory leak?

Where are the deletes? There are no deletes so there are memory leaks all
over the place.

The two pointers, r1 and r2, and array rr[] are local variables, when
the code exits function1(), these local variables should be released
automatically.
Am I right?

Wrong. The variables destructed and the memory they occupy is 'released',
BUT the memory they might be pointing to is not released.

void f()
{
X x;
X* xp = new X();
...
}

When you get to the end of this function, the memory for x and xp are both
released. That has nothing to do with the memory pointed to by xp, which is
a completely different thing. The memory for a pointer and the memory that
it points to are not the same thing.

It's very simple, every new must be matched by a delete.

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?

Thanks.

John (new to C++)

 

[JKop]

John posted:

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);

}

Oh Dear God No!


First thing, memorize the following:


Where I use "new", I must use "delete".

Where I use "new", I must use "delete".

Where I use "new", I must use "delete".

Where I use "new", I must use "delete".

Where I use "new", I must use "delete".

Where I use "new", I must use "delete".

Where I use "new", I must use "delete".




void f(X* pX0)
{
X* pX1; //I highly suggest that you prefix the letter
X* pX2; // p to the front of the names of pointer
X* pX3; //variables.

//We have just defined 3 variables of type X*
//These 3 variables, although being pointer variables,
//are just like any other variable! They have an
//address in memory and you can store a value in
//them. It just so happens that the value you store
//in them is a memory address!

//As you can see, I haven't initialized them,
//and therefore, they can contain any value!

pX1 = new X; //Here, I have set the value of the
//variable pX1.

//"new" is a sort-of function, what it does is
//allocate memory for an object of size sizeof(X) , and then
//it calls any constructors applicable to X .
//It then returns a value of type X* , having
//allocated the memory. You want to keep note of the
//address of this memory you've allocated, so you can
//actually access it in the future - and what better way
//and place to store it than in a pointer variable!

pX2 = new X;
pX3 = new X;

pX1 = pX0;
pX2 = pX0;
pX3 = pX0;

//You have just set the value of these 3 variables
//to some other value. One result is that now you
//no longer know the address of the memory you
//allocated using "new". Therefore, how the hell
//are you going to free that memory via a call to
//"delete"?, as you must specify the address of the
//previously allocated memory! Doh! And why is this so
//bad? Because the memory NEVER gets freed, even after
//you've exited your program. Then, if you run your
//program again, more memory is "leaked". If you
//continually do this, it *will* add up, especially on
//systems that are seldom turned off, eg. internet web
//servers.
}

main()
{
....
f(&X0); // I assume that you want to pass the address
// of a variable of type X and of name X0
....
f1(&X0);

}



-JKop

 

[John Harrison]

Does it mean the number of "new" is equal to the number of "delete" in
a code?

No, I mean that at run time every new must be matched by a delete. Often but
not always that means the news and delete are matched in code, but not
necessarily.

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?

Well I don't think you chosen a good example, because you have three memory
leaks in f.

Here's a simpler example

void delete_me(X* xp)
{
delete xp;
}

int main()
{
X* x1 = new X;
delete_me(x1);
X* x2 = new X;
delete_me(x2);
}

In the code there are two news and one delete. But at run time, new is
called twice and so is delete.

john

 

[Karl Heinz Buchegger]

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.

 

[Rolf Magnus]

You are right in that any and all constructors are not called yet, but
the actual memory *is* allocated for the variables at the beginning of
the code block... I think.

That depends entirely on the implementation. A conforming program cannot
find out what the implementation does.

 

[John]

Hi Karl:
Thank you very much. I understand now.
On this discussion list, there are so many warm-hearted experts who
spend their precious time instructing me--a newbie of C++.
I have two questions. Is the following code free of trouble?
void f()
{
X *x0 = new X(10);//create and initialize an object of class X.
X *x1;
x1 = x0;
delete x1; //release the memory that x0 uses.
}

I do not "delete" x0, but "delete" x1 to release the memory. Will it
cause any problem. An old code that I am modifying uses this approach.

Another question is:
What is the differece between the two lines below:
X *xx = new X;//line 1
X *xx1;//line 2
Line 1 declare xx and allocate it memory.
Line 2 declare xx1 and xx1 points to somewhere unknown. Has xx1 been
allocated memory?

Thanks a lot.

John