Does this cause a memory leak because pWinsock didn't get deleted inside the fonction?
Winsock* CreateWinsock()
{
Winsock* pWinsock=new Winsock;
return pWinsock;
}
Edit: Actually, I cannot delete my pointer since it is a member of Game (pWinsock) that received the newly created Winsock in the code above. Is there anything wrong with this?
class Game{
public:
Game();
~Game();
void CreateWindowClass(HINSTANCE);
void CreateRessources(HINSTANCE);
void ShowLoginScreen();
HWND Getm_hWnd();
public:
D2DResources* pD2DResources;
Winsock* pWinsock;
MessageLog* pMessageLog;
private:
HWND m_hWnd;
};
Watch out, if you delete the memory in the function the returned pointer will become a dangling pointer, as the memory for it has already been cleared. Dereferencing such a pointer is undefined behavior.
The program only causes a memory leak if the caller does not remember to delete the memory for themselves. Since you allocated memory within the function and returned it, it must be deleted somehow after the call. To delete the memory, it would look something like this:
Winsock *ptr = CreateWinsock(); // memory passed to ptr
// ...
delete ptr; // delete memory
The problem is that depending on the caller to delete the memory is quite cumbersome and unreliable. These kinds of potential problems can be alleviated through the use of smart pointers such as unique_ptr or shared_ptr. These objects delete the memory when their destructors are called, allowing great flexibility. Here's an example of how it would look for your program:
std::unique_ptr<Winsock> CreateWinsock()
{
return std::unique_ptr<Winsock>(new Winsock);
}
std::unique_ptr<Winsock> ptr = CreateWinsock();
There's no need to explicitly delete the pointer as the encapsulating smart pointer now has that responsibility.
If the caller of this function deletes the pointer after it uses it, there is no leak. Therefore, it is not appropriate to comment on the memory leak given just this piece of code.
To refrain from such a case where the caller forgets to delete the object, use shared pointer, or another smart pointer.
Nope, all that does is pass back the pointer to Winsock. For example
Winsock* ws = CreateWinsock();
ws->doSomething();
//......
//some more stuff
//......
//Finished using Winsock
delete ws;
If the delete wasn't called there when you're finished with Winsock then that would be seen as a memory leak because memory would be taken up by Winsock when its not being used anymore.
Related
always delete pointer even if it is just in function call stack?
Isn't it disappeared when function stack released?
// just Simple class
class CSimple{
int a;
}
// just simple function having pointer.
void simpleFunc(){
CSimple* cSimple = new CSimple();
cSimple->a = 10;
//.. do sth
delete cSimple; // <<< Always, do I have to delete 'cSimple' to prevent the leak of memory?
}
void main(){
for( int =0 ; i< 10 ; i++){
simpleFunc();
}
}
when function stack released?
It is true that "CSimple *csimple" goes away when the function returns.
However, there's a big difference between the pointer, and what it's pointed to.
When a pointer object gets destroyed, nothing happens to whatever the pointer is pointing to. There isn't just one, but two objects here:
The pointer.
What it's pointing to.
In this case, the pointer is pointing to an object in dynamic scope that was created with new.
Nothing is going to happen to this object, otherwise, so you will leak memory.
Therefore, this object needs to be deleted.
After you understand, and fully wrap your brain around this concept, your next step will be to open your C++ book to the chapter that talks about the std::unique_ptr and std::shared_ptr classes, which will take care of these pesky details, for you. You should learn how to use them. Modern C++ code rarely needs to delete something; rather these smart pointers do all the work.
Yes.
On scope exit (ex. when function exists or block { ... } finishes), all objects created on stack will be destroyed and memory will be freed.
This applies to your case, ie. the pointer will be destroyed and memory occupied by the pointer will be freed. The object pointed by the pointer will not be cleared.
This is a common problem and a nightmare when you deal with multiple flow paths (if-else ladders, many return statements) and exceptions.
To solve this problem, we employ 2 main strategies:
RAII
Smart pointers (std::unique_ptr, boost::scoped_ptr, legacy std::auto_ptr, etc).
RAII - without academic consideration - is just creating object on stack, like this:
{
std::string s;
fancy_object obj;
}
When we exit he scope, obj and s destructors will be called duing stack unwinding. Compiler ensures this for all flow paths and will keep proper order of deallocations for us.
If you need to allocate memory on heap, using new, use a smart pointer.
int foo()
{
std::unique_ptr<Object> o(new Object);
... some more code ...
if( something ) { return -1 }
... some more code ...
if( something_else ) { return -2 }
else { not yet }
return 0;
}
As you can see, we can leave the scope using 3 "exists". Normally, you'd need to clear your memory in all cases, which is prone to human errors.
Instead of clearing the object manually in all 3 palces, we rely on automatic destructor call for objects created on stack. Compiler will figure it out. When we leave the scope, std::unique_ptr destructor will be called, calling delete on Object.
Don't be affraid of smart poiners. They are not "slow", "bloat" or other nonsense. Smart poiners are designed to have no overhead on access, adding extra security.
Very similar technique is used for locks. Check out std::lock_guard class.
Yes, you must delete the data that is being pointed to.
The pointer itself is on the stack and does not need to be deleten.
You can, however, store cSimple on the stack, then you don't have to delete it:
void simpleFunc(){
CSimple cSimple; // no new
cSimple.a = 10; // "." instead of "->"
//.. do sth
// no deletion
}
I have the following below code snippet
class test{
private:
int *ptr;
public:
test(int *myptr){
ptr = myptr;
}
~test(){
delete ptr;
}
};
int main(){
int* myptr = new int;
*myptr = 10;
test obj(myptr);
delete myptr;
}
Is there a memory leak happening in this program, if I dont do a new to pointer in the class will space be allocated for that pointer?
Rule of thumb for dealing with allocating member: every new/new[] should be paired with exactly one delete/delete[]. If you are missing the delete, then you have a memory leak (you have allocated memory that you never clean up). If you are delete-ing multiples times, as you are doing in this code example, you will have memory corruption issues.
How are you delete-ing multiple times? In main(), you allocate a pointer:
int* myptr = new int;
You give a copy of that pointer to your test object, obj. Then, at the end of the scope we have:
{
// ...
delete myptr;
~test(); // implicit, which also does delete myptr
}
Only one of those two places should be responsible for delete-ing the pointer. Which one is based on the semantics of your code. Does test own the pointer? In which case, it should delete it but main() should not. Does it simply observe the pointer? In which case, it should not delete it. With C++11, we have new smart pointers to express this idea better.
I'd encourage you to browse the definitive C++ book list as concepts like this are very crucial to understanding C++ but also very difficult to explain properly in a short Q&A format.
You should only delte a pointer once in ~test() or delete directly
*** Error in `./a.out': double free or corruption (fasttop): 0x08d13a10 ***
Consider using std::shared_ptr<int> or std::unique_ptr<int>, since direct memory management is discouraged in modern C++ unless you've a reason to do so.
You can read about semantics differences of smart pointers here, and the reference for them is here.
It will crash when exit main function, a allocated in heap memory can't be release two times
What will happen in your instance is that the pointer will be given to delete in main(), then the same pointer value will be given a second time to delete in the destructor of obj. This is undefined behaviour, so it might work, or it might not, in any case, it isn't guaranteed to work, and therefore such a construct should not be used.
Ordinarily you would establish ownership rules, e.g. whether the constructor “takes ownership” (and is therefore responsible for freeing resources) is up to you, but typically, and especially in modern C++, ownership semantics can be clearly achieved by using std::unique_ptr and std::shared_ptr.
Most importantly, whichever method you use to determine ownership, make sure you are consistent! Avoid complicated ownership semantics, as it will make it much more difficult to detect memory-related issues (or more generally, resource-related issues).
Contrary to what other people have said: you do not delete pointers in C++, but objects.
What's the difference?
Let's simplify your code a bit:
int *myptr = new int; // 1
int *ptr = myptr; // 2
delete myptr; // 3
delete ptr; // 4
As far as the usage of new and delete is concerned, this is the same code - I've just removed the class, since it's not relevant to the point here.
This code does the following:
new int allocates an int, and returns its address. The address is then stored in myptr.
The address is copied to ptr. Both ptr and myptr contain the address of the int that was just allocated.
The int is deallocated.
The int is deallocated... but it was already deallocated? Oops!
If you're lucky, your program will crash at this point.
delete myptr; has nothing to do with the variable myptr, except that myptr holds the address of the thing to delete.
It doesn't even have to be a variable - you could do delete new int; (although it wouldn't be very useful) or delete someFunctionThatReturnsAnAddress();, or int *p = 1 + new int[2]; delete [] (p - 1);.
Your class has only a reference of the allocated integer. Consider to declare it int const* ptr, then you cant delete it inside the class.
You should remove delete ptr from the destructor.
Usually the scope which is allocating something is responsible for freeing it. In your case the main routine is allocating and has to free.
For your question "Is there a memory leak happening in this program", the answer is No, but an exception is rasing.
Your code logic is not good, you should delete the pointer at where it was created. In this example, you shouldn't delete ptr in destructor function.
How do I delete a pointer and the object it's pointing to?
Will the code below delete the object?
Object *apple;
apple = new Object();
delete apple;
And what happens if the pointer is not deleted, and gets out of scope?
Object *apple;
apple = new Object();
This might be a very basic question, but I'm coming from Java.
Your first code snippet does indeed delete the object. The pointer itself is a local variable allocated on the stack. It will be deallocated as soon as it goes out of scope.
That brings up the second point--if the pointer goes out of scope before you deallocate the object you allocated on the heap, you will never be able to deallocate it, and will have a memory leak.
Hope this helps.
Hello and welcome to C++ land! You will love how much you hate it (or something like that). C++, while appearing to be similar to java in untrained eyes might look similar, is actually quite different semantically. Lets see how these semantics play out in c++ in regards to your question. First lets take a class:
class Foo {
public:
Foo() { std::cout << "In constructor\n"; }
~Foo() { std::cout << "In destructor\n"; }
};
Foo here is just a representative of any class you might want to use. Lets see what happens when we create and play with a normal Foo object:
{
Foo bar;
do_stuff(bar);
}
If we were to run code that looked like this, we would see:
In constructor
In destructor
This is because, when an object is created, it is constructed using the constructor. When it goes out of scope, the destructor is called (~Foo in our code) which deconstructs (or destroys) the object. This is actually a fairly common and powerful feature in C++ (known as RAII, as opposed to other forms of returning memory to the system, such as Garbage Collection). Armed with this new knowledge, lets see what happens when we play with a pointer to Foo:
{
Foo *bar = new Foo();
some_more_stuff(bar);
}
What happens here is we would see:
In constructor
This is because of how pointers are allocated versus how variables are allocated. The way pointers are allocated, they don't actually go out of scope normally, but their contents do. This is known as a dangling pointer. For a better example, take a look at this:
#include <iostream>
int* get_int() {
int qux = 42;
int *foo = &qux;
return foo;
}
int main() {
int *qazal = get_int();
std::cout << *qazal;
}
Thanks to modern operating systems, this memory will still be returned when the program finishes, but not during the running of the program. If we were to delete the pointer (in the same scope it was created) via delete, then that memory will actually be returned to the operating system at that time.
When you call delete on a pointer it frees the memory of the thing pointed to. In other words you don't need to free the memory that makes up the pointer variable, just the thing that is pointed to. So your code:
Object *apple;
apple = new Object();
delete apple;
correctly deletes the object and there will be no memory leak.
If you don't call delete and the variable goes out of scope then you'll have a memory leak. These can be difficult to track down, so it's advisable to use a smart pointer class.
Operator delete deletes an object pointed to by the pointer that previously was allocated with operator new.
The pointer itself is not changed and even will have the same value as before the calling the operator. However its value becomes invalid after deleteing the object.
If the pointer itself is a local variable it will be destroyed after the control leaves the declaration region of the pointer.
If the pointer has the static storage duration then it will be destroyed when the program finishes its execution.
Take into account that you can use smart pointers instead of raw pointers.
For example
std::unique_ptr<Object> apple( new Object() );
in this case you need not to call delete It is the smart pointer that will do all the work itself.
Will the code below delete the object?
Yes, it will. But the pointer isn't deleted and accidentally using the pointer after deletion will lead to error.
And what happens if the pointer is not deleted, and gets out of scope?
Memory leak will happen.
I have a function foo like
myType** foo(){
myType **array = malloc( .... );
//Do some stuff
return array;
}
Here I have malloced , but did not free it as I am returning it . Will this cause a memory leak ? Should I explicitly free it in the calling function after I use it ?
It's a memory leak only if you don't free the memory (regardless where).
In this case, you should free it after the function is called and you're done with the pointer.
But this is the C way of doing it. In C++ you'd return a smart pointer, and use new instead of malloc.
With this type of function, the caller is the "owner" of the resources pointed at by the pointer. So the caller has to either free the resources, or pass them on to someone else who will.
In C++ one would favour returning a type that manages its own resources, for example an std::vector, or a smart pointer, that both takes care of resource de-allocation and makes the ownership clear.
See this example,and don't worry, read all about copy elision, particularly named return value optimization (NRVO).
std::vector<std::vector<SomeType>> foo()
{
std::vector<std::vector<SomeType>> tmp = ....;
// do some stuff
return tmp;
}
Memory leaks are caused by not freeing memory. As long as the memory will get freed at some point, then it isn't leaked. But if the pointer ever gets "lost" or if some sort of dependency loop causes a situation where the memory stays around even after it is no longer useful, then at that point you've created a leak.
Usually in this type of situation, you create a pattern like this:
void* mything_init() {
void *obj = malloc(LEN);
// do some initialization
return obj;
}
void mything_uninit(void* thing) {
// any teardown
free(thing);
}
For every mything that you _init, you must eventually call _uninit on that object. That is your responsibility as the user of that library. While as the writer of that library, you make sure that anything you allocate in the _init gets properly freed in _uninit.
This is a C pattern rather than a C++ pattern as malloc() is a C idiom. C++ uses new and delete, and this sort of work us usually done in a constructor/destructor pair.
I had issues about this problem for long time. I suggest you to read this wiki page:
Smart Pointers
In your case you can use unique_ptr. So you would be able to transfer the ownership of the pointer to the caller function.
unique_ptr<int> callee1()
{
std::unique_ptr<int> p1(new int(5));
return p1;
}
int main()
{
std::unique_ptr<int> result = callee1() ;
(*result)++; // Increase the value stored in the pointer
std::cout << "New Value: " << (*result) << std::endl;
result.reset() ;
}
Hope this code sniper helps you.
I have a class pointer declaration:
MyClass* a;
In destruction method I have:
if (a)
{
delete a;
a= NULL;
}
I got a problem when delete the pointer a:
Attempted to read or write protected memory. This is often an indication that other memory is corrupt.
What is the cause of the problem and how can I get rid of it?
With your current declaration:
MyClass* a;
a gets a random value. If you never give it a valid value later, such as:
a = new MyClass();
It will point to an unknown place in memory, quite probably not a memory area reserved for your program, and hence the error when you try to delete it.
The easiest way to avoid this problem is to give a a value when you declare it:
MyClass* a = new MyClass();
or, if you cannot give it a value when you declare it (maybe you don't know it yet), assign it to null:
MyClass* a = 0;
By the way, you can remove the test (if (a)) from your code. delete is a no-op on a null pointer.
Use smart pointer to free memory. delete in application code is always wrong.
unless you have initialized the pointer to something after this:
MyClass* a;
the pointer a will hold some random value. So your test
if (a) { }
will pass, and you attempt to delete some random memory location.
You can avoid this by initializing the pointer:
MyClass* a = 0;
Other options are that the object pointed to has been deleted elsewhere and the pointer not set to 0, or that it points to an object that is allocated on the stack.
As has been pointed out elsewhere, you could avoid all this trouble by using a smart pointer as opposed to a bare pointer in the first place. I would suggest having a look at std::unique_ptr.
How did you allocate the memory that a points to? If you used new[] (in order to create an array of MyClass), you must deallocate it with delete[] a;. If you allocated it with malloc() (which is probably a bad idea when working with classes), you must deallocate it with free().
If you allocated the memory with new, you have probably made a memory management error somewhere else - for instance, you might already have deallocated a, or you have written outside the bounds of some array. Try using Valgrind to debug memory problems.
You should use
MyClass* a = NULL;
in your declaration. If you never instantiate a, the pointer is pointing to an undefined region of memory. When the containing class destructor executes, it tries to delete that random location.
When you do MyClass* a; you declare a pointer without allocating any memory. You don't initialize it, and a is not necessarily NULL. So when you try to delete it, your test if (a) succeeds, but deallocation fails.
You should do MyClass* a = NULL; or MyClass* a(nullptr); if you can use C++11.
(I assume here you don't use new anywhere in this case, since you tell us that you only declare a pointer.)