Functions like wcsdup, implicitly calls malloc to allocate memory for the destination buffer. I was wondering as the memory allocation is not very explicit, so does it seems logical to explicitly free the storage?
This is more like a design dilemma and the reasons for and against are as follows
Should be freed because
Not freeing it would cause Memory Leak.
It is well documented that wcsdup/_wcsdup calls malloc to allocate memory even when its called from a C++ Program.
Should not be freed because
Memory accumulated by wcsdup would eventually be freed when program exits. We always live with some memory leaks through out the program lifetime(Unless we are heavily calling wcsdup for large buffer size).
It can be confusing as free was not preceded by an explicit malloc.
As its not part of the standard but posix compliant, Microsoft implementation may not use malloc for allocating destination buffer.
What should be the approach?
From MSDN:
it is good practice always to release this memory by calling the free routine on the pointer returned
From the page you linked:
The returned pointer can be passed to free()
It seems fairly explicit: if you care about memory leaks, then you should free the memory by using free.
To be honest, I'm concerned about the cavalier attitude hinted at with this:
We always live with some memory leaks through out the program lifetime
There are very rarely good reasons to leak memory. Even if the code you write today is a one-off, and it's not a long-lived process, can you be sure that someone's not going to copy-and-paste it into some other program?
Yes, you should always free heap-allocated memory when you're done using it and know that it is safe to do so. The documentation you link to even states:
For functions that allocate memory as if by malloc(), the application
should release such memory when it is no longer required by a call to
free(). For wcsdup(), this is the return value.
If you are concerned about the free being potentially confusing, leave a comment explaining it. To be honest, though, that seems superfluous; it's pretty obvious when a pointer is explicitly freed that it's "owned" by the code freeing it, and anyone who does become confused can easily look up the wcsdup documentation.
Also, you should really never have memory leaks in your program. In practice some programs do have memory leaks, but that doesn't mean it's okay for them to exist. Also note that just because you have a block of memory allocated for the entire lifespan of the program, it is not leaked memory if you are still using it for that entire duration.
From your own link:
For functions that allocate memory as if by malloc(), the application should release such memory when it is no longer required by a call to free().
From MSDN:
The _strdup function calls malloc to allocate storage space for a copy of strSource and then copies strSource to the allocated space.
and strdup is deprecated as from MSVC 2005 and calling it calls _strdup so it is using malloc
Related
I am getting some sort of memory leak with this code. Although is doesnt actually throw any error, when run through Memcheck, it detects a memory leak.
Am I supposed to also delete[] the temporary array or something?
Here is the function it is happening in:
Every memory allocation needs to be paired to exactly one deallocation. When this function ends, the memory pointed to by newBinaryNum has not been deallocated, and the address of that memory is not retained anywhere. Hence, memory leak.
Since you don't use any benefits of dynamic allocation for this BinaryNum object, don't use dynamic allocation. Just declare your variable as BinaryNum newBinaryNum;. Not only will the compiler handle the memory needs of this object, but also your code would be able to benefit from return value optimization.
(Also, you should lean on the standard library a bit more. Your manipulations of binaryAry mimic the functionality of std::vector. Why re-invent the wheel? Your focus in this function should be on the BinaryNum class, so delegate the memory management details to the standard library.)
This question is as in title:
Is it possible to produce a memory leak without using any kernel specific means like malloc, new, etc?
What if I will make a linked list inside a function with lot of elements in there, and after it I'll exit from this function without cleaning a list. The list will be created without using any malloc calls, i.e.
struct list_head {
struct list_head *next, *prev;
}
Can it be guaranteed that all resources will be freed after exiting from this function? So I can freely execute it a million times and nothing will be leaked?
Subject: If you not using any particular malloc or new calls you won't get a heap memory leak. Never. Is that right?
A leak is always connected to a resource. A resource is by definition something that you acquire manually, and that you must release manually. Memory is a prime example, but there are other resources, too (file handles, mutex locks, network connections, etc.).
A leak occurs when you acquire a resource, but subsequently lose the handle to the resource so that nobody can release it. A lesser version of a leak is a "still-reachable" kind of situation where you don't release the resource, but you still have the handle and could release it. That's mostly down to laziness, but a leak by contrast is always a programming error.
Since your code never acquires any resources, it also cannot have any leaks.
The variables you applied without malloc or new is located at stack
space in the memory. So when the function returned, the variable is
taken back.
On the other hand, the memory you applied with malloc or new is
located at heap space. The system doesn't care whether you release the
space or not. In this situation, if you don't use free or delete,
memory leak will happen.
Subject: If you not using any particular malloc or new calls you won't get a heap memory leak. Never. Is that right?
That assumption is not entirely correct. The problem is that the operating system itself (or other third party components you have to rely on) can have memory leaks as well. In that case you might not actively call malloc, but call other (operating system) functions which could leak.
So your assumption depends on how strongly you consider such a thing. You can argue that the OS/third party implementation is outside your domain, then this assumption would be correct. If you have a well defined system and your requirements are such that you have to garuantee a certain uptime, something like this may have to be considered as well.
So the answer to this question ...
Is it possible to make memory leak without using malloc?
... is:
Yes, it is possible.
malloc() allocates memory from the heap, while space for string and struct literals (string1, string2 and those list_head's) will be reserved at compile time at the stack.
Actually any memory allocated for a program (heap or stack) will be reclaimed by the kernel when the process exits (at *nix system at least).
I would define memory leak as allocating memory on heap and without freeing it when your program exits. This definition actually answers your question.
There are standard functions (like strdup) that will allocate memory on heap, beware of them.
Another example of a resource that you can allocate and forget to free:
If you're using OpenGL, and you call glGenBuffers() a million times without the corresponding glDeleteBuffers calls, it's extremely likely that you will run out of VRAM and your graphics driver will start leaking to system memory.
I just had this happen. Fortunately, Visual Studio's memory profiler made it pretty easy to find. It showed up as a large number of allocations made by the external process nvoglv32.dll, which is my NVIDIA OpenGL driver.
Using c:
char ptr[n];
free(ptr);
In my opinion: when "char ptr[n];" is used, the memory is allocated, and ptr is pointed to it, free(ptr) should work.
And the program failed, why?(n == 5 e.g.)
Any deep analysis?
Because you called free on a variable not allocated with malloc.
This causes Undefined Behavior. Luckily for you it crashes and you can detect it, else it can crash at most awkward times.
You call free for deallocating memory of heap allocated variables, What you have is an array on local storage(assuming it to be in a function) and it automatically deallocates when the scope({,}) in which it was created ends.
Because this is undefined behavior what you're doing. (It means it can literally can do anything, including crashing, running seemingly fine, making daemons fly out of your nose, etc.) You can only free() a pointer that you acquired using malloc().
Auto arrays do not have to be free()'d. They are deallocated when their scope ends.
Only free an object that has been allocated by malloc. Freeing an object that has not been allocated by malloc is undefined behavior.
Because of `char ptr[n];' is away to declare an array in STACK memory, and it has scope of the block, which mean it destroyed from the memory when the block is finish.
but when you use malloc(size) the pointer will point to in a piece of memory in the HEAP memory and it take the scope which the programer give it. I mean that when you want to destroy it you must use free(ptr) or OS will free it after the program finish.
So, when you use free on pointer that point to a piece of memory in STACK memory it cause Undefined Behavior and the program crash, because free operates only on the HEAP memory.
This looks similar can a call to free in c ever fail(SO)
The behavior is undefined as per the standard, In some cases your code would not crash so soon. It may corrupt the heap and crash very late during execution and make the debugging kind of difficult.
In a way it depends on the design of malloc/free methods.
One way which I know is :
with each malloc, an extra block of memory is attached to the block which is returned by malloc(). This block contains some housekeeping data which is needed while a call to free(). In your case this data is missing since the memory was not allocated by malloc(). So free()
is trying to use the data preceding your array without knowing that its junk.
I am here stuck with a question in my C++ book with the following:
"What does the use of new require you to also call delete?"
Maybe you have an answer for that?
Because that is the way C++ is designed & that is the intended behavior.
The intention was to provide a memory allocation which you demand and own till you reliquish it explicitly.
new gives you a dynamic memory allocation(on heap) which will continue to exist and you own it untill you explicitly deallocate it by calling delete.
Failing to call a delete on a newed buffer will lead to Undefined Behaviors usually in the form of. 1 memory leaks.
1 This was discussed here.
when you do a new, OS allocates the memory to the pointer you are assigning it. After your usage is completed you may not require it anymore. But the memory is still marked as "being used" by OS.
Now, when the pointer is declared in a scope of a function or any other block (of {}), it will be deleted (only pointer will be removed) when the execution of the block is over. In such cases the memory that was allocated using new is remained marked "being used" by OS and is not allocated to any other pointer that calls new or to a variable. This causes an orphan block of memory in RAM, that will never be used because its pointer was removed from memory but it will occupy a memory block.
This is called a memory leak. A few of such blocks may make your application unstable as well.
You use delete to free such memory blocks and relieve the OS so that it can be used well for other requests
There is no Garbage Collector in C++, and therefore you are responsible for deallocating the allocated memory. Anyway, the operating system "knows" what memory your program allocated. So when your program exits, the operating system is again responsible for the memory. But if you have a long running C++ program and never call delete noone will help you to get rid of your garbage.
Calling new has allocated memory for the object and it has also arranged for the constructor of that object to be executed.
You could free the memory by calling free(), but you should actually use delete to free memory allocated by new, since this will also cause the objects destructor to be executed.
In C++, when you make a new variable on the heap like this:
int* a = new int;
you can tell C++ to reclaim the memory by using delete like this:
delete a;
However, when your program closes, does it automatically free the memory that was allocated with new?
Yes, it is automatically reclaimed, but if you intend to write a huge program that makes use of the heap extensively and not call delete anywhere, you are bound to run out of heap memory quickly, which will crash your program.
Therefore, it is a must to carefully manage your memory and free dynamically allocated data with a matching delete for every new (or delete [] if using new []), as soon as you no longer require the said variable.
When the process is terminated the memory is reclaimed back by the OS. Of course this argument shouldn't in any case be used to not perform proper memory management by the program.
Don't let people tell you yes. C++ has no concept of an OS, so to say "yes the OS will clean it up" is no longer talking about C++ but about C++ running on some environment, which may not be yours.
That is, if you dynamically allocate something but never free it you've leaked. It can only end its lifetime once you call delete/delete[] on it. On some OS's (and almost all desktop OS's), memory will be reclaimed (so other programs may use it.) But memory is not the same as resource! The OS can free all the memory it wants, if you have some socket connection to close, some file to finish writing to, etc, the OS might not do it. It's important not to let resources leak. I've heard of some embedded platforms that won't even reclaim the memory you've not freed, resulting in a leak until the platform is reset.
Instead of dynamically allocating things raw (meaning you're the one that has to explicitly delete it), wrap them into automatically allocated (stack allocated) containers; not doing so is considered bad practice, and makes your code extremely messy.
So don't use new T[N], use std::vector<T> v(N);. The latter won't let a resource leak occur. Don't use new T;, use smart_ptr p(new T);. The smart pointer will track the object and delete it when it's know longer used. This is called Scope-bound Resource Management (SBRM, also known as the dumber name Resource-Acquisition is Initialization, or RAII.)
Note there is no single "smart_ptr". You have to pick which one is best. The current standard includes std::auto_ptr, but it's quite unwieldy. (It cannot be used in standard containers.) Your best bet is to use the smart pointers part of Boost, or TR1 if your compiler supports it. Then you get shared_ptr, arguably the most useful smart pointer, but there are many others.
If every pointer to dynamically allocated memory is in an object that will destruct (i.e., not another object that is dynamically allocated), and that object knows to free the memory, that pointer is guaranteed to be freed. This question shouldn't even be a problem, since you should never be in a position to leak.
No, when the program exits ("closes") the dynamically allocated memory is left as is
EDIT:
Reading the other answers, I should be more precise. The destructors of dynamically allocated objects will not run but the memory will be reclaimed anyway by any decent OS.
PS: The first line should read
int* a = new int;
No, it's your responsibility to free it. Also, a must be a pointer, so it should be:
int *a = new int;
delete a;
This excellent answer by Brian R. Bondy details why it's good practice to free the memory allocated by a.
It is important to explicitly call
delete because you may have some code
in the destructor that you want to
execute. Like maybe writing some data
to a log file. If you let the OS free
your memory for you, your code in your
destructor will not be executed.
Most operating systems will deallocate
the memory when your program ends. But
it is good practice to deallocate it
yourself and like I said above the OS
won't call your destructor.
As for calling delete in general, yes
you always want to call delete, or
else you will have a memory leak in
your program, which will lead to new
allocations failing.
When your process terminates, the OS does regain control of all resources the process was using, including memory. However, that, of course, will not cause C++'s destructors to be necessarily run, so it's not a panacea for not explicitly freeing said resources (though it won't be a problem for int or other types with noop dtors, of course;-).