When malloc is called, the size is stored adjacent to the allocated block so that free will know how much to free etc ( http://c-faq.com/malloc/freesize.html ).
My question is, Say we have dynamically allocated memory and later in the code we increment the pointer
pointer++
And then later, if i call a
free(pointer)
what memory does actually get freed up.
Is it number of allocated bytes starting from the current address pointed by 'pointer' or from the base address to which it has been allocated.
You need to free() the same pointer as you received from malloc(). Incrementing, altering or changing it is undefined behaviour, that is usually a segmentation fault.
Think of the pointer you receive as a book from a library. You get home and read it. Afterwards you remove the front page and the book's back and hand it back to the librarian. Will he accept it or are you in serious trouble now? ;-)
You can only call free() on a value that you previously obtained from malloc(), calloc(), or realloc() (or NULL). Everything else is undefined.
For example, an implementation might store the size of the allocated block in 4 bytes before the return address from malloc(). Then, free() goes back 4 bytes and finds out the size. This wouldn't work if you don't pass the original pointer back to free().
It will cause undefined behavior. Most likely it will crash your program either instantly or later.
That's undefined behavior. And it will most probably results in problem later on.
If you increment the pointer without saving the original malloced location you can't call free on it. You have to save the original location somewhere and use a temporary point when you increment.
doing pointer++ to original pointer is terribly wrong. result of freeing it may be different on different implementations, but you definitely shouldn't do it.
The code managing the free storage just assumes that you wouldn't hand it the wrong pointer. It takes whatever you give, doesn't check its plausibility, and interprets it the same way it would interpret the right pointer. It will act according to whatever values it reads from whatever memory locations it looks at assuming the pointer was rightfully obtained. If you handed it a stray pointer, it will find nonsensical values and thus act nonsensical.
This is called undefined behavior and it's a mean thing. It might format your hard drive, toast your CPU, or make your program seemingly work the way it is expected to until you retire. You never know.
This is what we call a memory leak/segmentation fault.
You HAVE to pass the same pointervalue to free() as the one you got from malloc() or your application will misbehave/crash.
The pointer returned by malloc() points directly to the memory on the heap that will be used by your program.
However, this isn't the only memory that's allocated. A few bytes are allocated in the memory locations immediately preceding the pointer returned that indicate the size of the chunk on the heap. This isn't used by your program, but it will definitely be needed by free.
When free(p) is called, the information about its chunk on the heap is contained in, say, the locations from p-4 through p-1. This depends on implementation of course, but the details need not concern the programmer. The only thing that the programmer needs to know is that free uses that area of memory to free the chunk of memory from the heap, and that area is derived from the original pointer p.
In other words, if you call free on p, it will only make sense if malloc once returned exactly p.
If you pass in a pointer that wasn't created with malloc, who knows what will lie at p-1, p-2, etc.? It will probably result in a catastrophic failure.
Related
I'm trying to figure out how many bytes in a block are taken up by the boundary tags. I have been told that when trying to malloc an adjacent block of memory, a "jump" will appear in assembly code, and I can use that to determine the size of the boundary tag. I've tried this:
int* arr = malloc(8);
arr++;
arr = malloc(8);
But there isn't any jump in assembly code. Am I "trying to malloc an adjacent block of memory"?
EDIT: I think he means a jump will appear between address value. I use the beginning of the second block of memory subtract the payload size of the first block. But I'm still confused, how could I malloc an adjacent block of memory?
Unless you're writing an actual memory allocator, you can't actually allocate two consecutive chunks of memory. If you want to see some pretty gnarly code which does this, have a look at the Illumos malloc https://github.com/illumos/illumos-gate/blob/master/usr/src/lib/libc/port/gen/malloc.c.
If you want to see how Illumos (and Solaris) handle the redzone between allocated blocks, you should trawl through https://github.com/illumos/illumos-gate/tree/master/usr/src/lib/libumem/common.
The memory consumed by malloc(3) requires, for proper management of the actually used memory, of some structures that must be dynamically allocated also. For this reason, many allocators just do allocate the space required for the management data adjacent to the block space dedicated to the user. This makes that normally two consecutive junks of memory allocated by malloc(2) show some gap in their addresses.
There are other reasons to see gaps, one fundamental is that malloc normally gives you aligned memory addresses, so it is warranted that the data you store on that memory will be properly aligned.
And of course, there can be implementations (normally when heap allocation should be more robust in respect to buffer overruns) that the memory dedicated to storage of management data is completely unrelated and apart off the final given memory. In this case you could observe no gaps between memory allocations on some cases.
Anyway, your code has serious bugs, let's see:
int* arr = malloc(8);
You had better here to acquire just the memory you need, using the sizeof operator, as in int *arr = malloc(sizeof *arr); instead.
arr++;
this statement is useless, as you are going to overwrite the value of arr (the pointer) with new assignment statement after it from malloc(), so it is of no use to increment the pointer value. You also are somewhat losing the returned value of the previous malloc() (which is essential in order to return the allocated memory, but read below).
arr = malloc(8);
Until here, you had the chance to --arr decrementing the value of arr in order to be capable of free(3) that block. But this statement overwrites the value stored in arr so the previous pointer value is overwritten by the new pointer. Memory you acquired on the first malloc has no way to be accessed again. This is what is commonly known as a memory leak, and is normally a serious error (very difficult to catch) on long run programs (like servers or system daemons). The program allocates a bunch of memory in the inner part of a loop, that is not returned back with a call to free(3), so the program begins growing and growing until it exhausts all the available memory.
A final note, I don't understand what did you mean with malloc adjacent block of memory. Did you believe that incrementing the pointer would make malloc() to give you a special block of memory?
First, malloc has no idea of what are you going to do with the pointer it gives to you.
But also, it doesn't know anything about the variable contents of the pointer you are assigning to (you can even not store it in a variable, and pass it as another parameter to another functions) So there's no possibility for malloc to know that you have incremented the pointer value, or even the pointer location, from its body.
I cannot guess your interpretation. It would be nice to know what has made you to think that you can control how malloc(3) selects the block of memory to give to you. You have no control on the internals of malloc() You just specify the amount of continous memory you want, and mallocs provides it, giving you a pointer pointing to the start of that block. You cannot assume that the next time you call malloc (with the same or different amount of memory) it will give you an adjacent block. It just can be completely unrelated (above or below in memory) to the previous given block. And you cannot modify that pointer, because you need it to call free(3) once you don't need the block anymore, with exactly the same pointer value that malloc(3) gave to you. If, for some reason you modify the pointer, you need to restore it to the original value to be capable of calling free(3). Lack to do so, you'll probably crash your program at the next call to free(3).
I just see a memory leak. Malloc 2 times into different vars 8 bytes of space and see if the difference is more than 8 bytes or 2 int.
I am Confused and thinking about Free Pointer in Following way , can someone clarify
I always thought that the free function just removes the address stored in the pointer, hence breaking the connection between the pointer and the space allocated for block. At the end making the values at the space a garbage value and hence making it virtually free to use by another program.
Please specify the reason why I am wrong.
Also, What happens when you point two pointers to a single space allocated
and Try to free them both? Does it Gives a compilation Error / Run Time Error?
free(ptr) de-allocates the memory block pointed to by its argument ptr, that is it returns the block to the free pool. Consecutive calls to malloc() may use that memory, re-allocating it for another purpose.
However, free() receives its argument as a copy, that is it gets the value from the ptr variable, not the variable itself. As a result it can't 'clear the pointer'. If you expect the ptr variable becomes NULL, you're wrong. That's a common mistake. You should assign ptr = NULL yourself after free() to make sure the variable value becomes unusable.
As for the memory block itself: free() is not obliged to 'make it garbage' or clear it. You can try accessing the freed memory by the same pointer, and often you will see the previous content still present there.
But don't rely on it! The memory can be reallocated for another purpose almost any time (the freed memory might even become unavailable for the process), so fiddling with it may invoke unpredictable results (so-called Undefined Behavior).
An attempt to deallocate a block twice (i.e. a second free() with the same pointer) is UB, too.
I have been taught in lectures, that calling free() on a pointer twice is really, really bad. I know that it is good practice, to set a pointer to NULL, right after having freed it.
However, I still have never heard any explanation as to why that is. From what I understand, the way malloc() works, it should technically keep track of the pointers it has allocated and given you to use. So why does it not know, whether a pointer it receives through free() has been freed yet or not?
I would love to understand, what happens internally, when you call free() on a location that has previously already been freed.
When you use malloc you are telling the PC that you want to reserve some memory location on the heap just for you. The computer gives back a pointer to the first byte of the addressed space.
When you use free you are actually telling the computer that you don't need that space anymore, so it marks that space as available for other data.
The pointer still points to that memory address. At this point that same space in the heap can be returned by another malloc call. When you invoke free a second time, you are not freeing the previous data, but the new data, and this may not be good for your program ;)
To answer your first question,
So why does it not know, whether a pointer it receives through free() has been freed yet or not?
because, the specification for malloc() in C standard does not mandate this. When you call malloc() or family of functions, what it does is to return you a pointer and internally it stores the size of the memory location allocated in that pointer. That is the reason free() does not need a size to clean up the memory.
Also, once free()-d, what happens with the actually allocated memory is still implelentation dependent. Calling free() is just a marker to point out that the allocated memory is no longer in use by the process and can be reclaimed and e re-allocated, if needed. So, keeping track of the allocated pointer is very needless at that point. It will be an unnecessary burden on the OS to keep all the backtracks.
For debugging purpose, however, some library implementations can do this job for you, like DUMA or dmalloc and last but not the least, memcheck tool from Valgrind.
Now, technically, the C standard does not specify any behaviour if you call free() on an already free-ed pointer. It is undefined behavior.
C11, chapter ยง7.22.3.3, free() function
[...] if
the argument does not match a pointer earlier returned by a memory management
function, or if the space has been deallocated by a call to free() or realloc(), the
behavior is undefined.
C standard only says that calling free twice on a pointer returned by malloc and its family function invoke undefined behavior. There is no further explanation why it is so.
But, why it is bad is explained here:
Freeing The Same Chunk Twice
To understand what this kind of error might cause, we should remember how the memory manager normally works. Often, it stores the size of the allocated chunk right before the chunk itself in memory. If we freed the memory, this memory chunk might have been allocated again by another malloc() request, and thus this double-free will actually free the wrong memory chunk - causing us to have a dangling pointer somewhere else in our application. Such bugs tend to show themselves much later than the place in the code where they occured. Sometimes we don't see them at all, but they still lurk around, waiting for an opportunity to rear their ugly heads.
Another problem that might occure, is that this double-free will be done after the freed chunk was merged together with neighbouring free chunks to form a larger free chunk, and then the larger chunk was re-allocated. In such a case, when we try to free() our chunk for the 2nd time, we'll actually free only part of the memory chunk that the application is currently using. This will cause even more unexpected problems.
When you are calling malloc you are getting a pointer. The runtime library needs to keep track of the malloced memory. Typically malloc does not store the memory management structures separated from the malloc ed memory but in one place. So a malloc for x bytes in fact takes x+n bytes, where one possible layout is that the first n bytes are containing a linked list struct with pointers to the next (and maybe previous) allocated memory block.
When you free a pointer then the function free could walk through it's internal memory management structures and check if the pointer you pass in is a valid pointer that was malloced. Only then it could access the hidden parts of the memory block. But doing this check would be very time consuming, especially if you allocate a lot. So free simply assumes that you pass in a valid pointer. That means it directly access the hidden parts of the memory block and assumes that the linked list pointers there are valid.
If you free a block twice then you might have the problem that someone did a new malloc, got the memory you just freed, overwrites it and the second free reads invalid pointers from it.
Setting a freed pointer to NULL is good practice because it helps debugging. If you access freed memory your program might crash, but it might also just read suspicious values and maybe crash later. Finding the root cause then might be hard. If you set freed pointers to NULL your program will immediately crash when you try to access the memory. That helps massively during debugging.
I have been recently trying to learn how to program in the C programming language.
I am currently having trouble understanding how memory is deallocated by free() in C.
What does it mean to free or release the memory?
For instance, if I have the following pointer:
int *p = malloc(sizeof(int));
When I deallocate it using free(p), what does it do? Does it somehow flag it as "deallocated", so the application may use it for new allocations?
Does it deallocates only the pointer address, or the address being pointed is also deallocated too?
I would do some experiments myself to better understand this, but I am so newbie in the subject that I don't know even how to debug a C program yet (I'm not using any IDE).
Also, what if int *p is actually a pointer to an array of int?
If I call free(p), does it deallocate the whole array or only the element it is pointing to?
I'm so eager to finally understand this, I would very much appreciate any help!
What does it mean to free or release the memory?
It means that you're done with the memory and are ready to give it back to the memory allocator.
When I deallocate it using free(p), what does it do?
The specifics are implementation dependent, but for a typical allocator it puts the block back on the free list. The allocator maintains a list of blocks that are available for use. When you ask for a chunk of memory (by calling malloc() or similar) the allocator finds an appropriate block in the list of free blocks, removes it (so it's no longer available), and gives you a pointer to the block. When you call free(), the process is reversed -- the block is put back on the free list and thereby becomes available to be allocated again.
Importantly, once you call free() on a pointer, you must not dereference that pointer again. A common source of memory-related errors is using a pointer after it has been freed. For that reason, some consider it a helpful practice to set a pointer to nil immediately after freeing it. Similarly, you should avoid calling free() on a pointer that you didn't originally get from the allocator (e.g. don't free a pointer to a local variable), and it's never a good idea to call free() twice on the same pointer.
Does it deallocates only the pointer address, or the address being pointed is also deallocated too?
When you request a block of memory from the allocator, you specify the size of the block you want. The allocator keeps track of the size of the block so that when you free the block, it knows both the starting address and the block size. When you call free(p), the block that p points to is deallocated; nothing happens to the pointer p itself.
Also, what if int *p is actually a pointer to an array of int?
An array in C is a contiguous block of memory, so a pointer to the first element of the array is also a pointer to the entire block. Freeing that block will properly deallocate the entire array.
I'm so eager to finally understand this, I would very much appreciate any help!
There are a number of good pages about memory allocation in C that you should read for a much more detailed understanding. One place you could start is with the GNU C Library manual section on memory allocation.
As alluded to above and in the other answers, the actual behavior of the allocator depends on the implementation. Your code shouldn't have any particular expectations about how memory allocation works beyond what's documented in the standard library, i.e. call malloc(), calloc(), etc. to get a block of memory, and call free() to give it back when you're done so that it can be reused.
malloc and free do whatever they want. Their expected behaviour is that malloc allocates a block of desired size in dynamic memory and returns a pointer to it. free must be able to receive one such pointer and correctly deallocate the block. How they keep track of the block size is irrelevant.
Is int *p a pointer to an array of ints ? Maybe. If you allocated sufficient space for several ints, yes.
There is a fixed and limited amount of memory in your computer, and everybody wants some. The Operating system is charged with the task of assigning ownership to pieces of memory and keeping track of it all to assure that no one messes with anyone else's.
When you ask for memory with malloc(), you're asking the system (the C runtime and the OS) to give you the address of a block of memory that is now yours. You are free to write to it and read from it at will, and the system promises that no one else will mess with it while you own it. When you de-allocate it with free(), nothing happens to the memory itself, it's just no longer yours. What happens to it is none of your business. The system may keep it around for future allocations, it may give it to some other process.
The details of how this happens vary from one system to another, but they really don't concern the programmer (unless you're the one writing the code for malloc/free). Just use the memory while it's yours, and keep your hands off while it's not.
I have allocated memory to some data type and assigned some value. Now using free is the data in the memory deleted or not? What is use of using free if the data assigned is not deleted? Can anyone help me out? Ex:
int *arr;
arr=(int*)malloc(sizeof(int)*1000);
assert(arr!=NULL);
/*Some operation*/
arr[123]=354;
//some operations
printf("%d",*(arr+123));
//calling some funcs
free(arr);
printf("\n%d",*(arr+123));
The point of free is to make the memory you allocated available for following calls to malloc. It does not guarantee that the buffer passed to it is wiped in any way.
In fact, what you're doing is provoking undefined behavior; accessing a buffer that has been free'd might give the value that was previously stored in it, or any other value, or it might crash your program, or do anything else.
You cannot "delete" things from memory.
What free() does is it reclaims the memory, so that it can be recycled by a future call to malloc().
You cannot legally dereference a pointer that you got from malloc() (or any other dynamic allocation call) after calling free() on it. Doing so invokes undefined behavior.
By using malloc you set aside a specified number of bytes to be used for what ever you want. Returned is an address to the area. The memory area is also flagged in a special table as occupied meaning it can't be reserved by another process or another malloc.
When you free the allocated memory the area is flagged as free, but the data is not erased. If you want to erase the data you'll have to do a memset or manually loop and set data to 0 or some other value before you free it.
As mentioned by others the operation on free memory is undefined.
A short introduction to how memory can be arranged. Look i.e. at heap:
http://www.geeksforgeeks.org/archives/14268 or Fig 1. at bottom of this comment: Strange global variable behaviour, once variable name is changed issue disappears
I have written some more about the malloc process itself here:
Can some explain the performance behavior of the following memory allocating C program? where also virtual vs. physical memory, pages etc are some of the points.