I am struggling to wrap my head around malloc in c - specifically when it needs to be free()'d. I am getting weird errors in gcc such as:
... free(): invalid next size (fast): ...
when I try to free a char pointer. For example, when reading from an input file, it will crash on certain lines when doing the following:
FILE *f = fopen(file,"r");
char x[256];
while(1) {
if(fgets(x,sizeof x,f)==NULL) break;
char *tmp = some_function_return_char_pointer(x); //OR malloc(nbytes);
// do some stuff
free(tmp); // this is where I get the error, but only sometimes
}
I checked for obvious things, such as x being NULL, but it's not; it just crashes on random lines.
But my REAL question is - when do I need to use free()? Or, probably more correctly, when should I NOT use free? What if malloc is in a function, and I return the var that used malloc()? What about in a for or while loop? Does malloc-ing for an array of struct have the same rules as for a string/char pointer?
I gather from the errors I'm getting in gcc on program crash that I'm just not understanding malloc and free. I've spent my quality time with Google and I'm still hitting brick walls. Are there any good resources you've found? Everything I see says that whenever I use malloc I need to use free. But then I try that and my program crashes. So maybe it's different based on a variable's scope? Does C free the memory at the end of a loop when a variable is declared inside of it? At the end of a function?
So:
for(i=0;i<100;i++) char *x=malloc(n); // no need to use free(x)?
but:
char *x;
for(i=0;i<100;i++) {
x=malloc(n);
free(x); //must do this, since scope of x greater than loop?
}
Is that right?
Hopefully I'm making sense...
malloc() is C's dynamic allocator. You have to understand the difference between automatic (scoped) and dynamic (manual) variables.
Automatic variables live for the duration of their scope. They're the ones you declare without any decoration: int x;
Most variables in a C program should be automatic, since they are local to some piece of code (e.g. a function, or a loop), and they communicate via function calls and return values.
The only time you need dynamic allocation is when you have some data that needs to outlive any given scope. Such data must be allocated dynamically, and eventually freed when it is no longer necessary.
The prime usage example for this is your typical linked list. The list nodes cannot possibly be local to any scope if you are going to have generic "insert/erase/find" list manipulation functions. Thus, each node must be allocated dynamically, and the list manipulation functions must ensure that they free those nodes that are no longer part of the list.
In summary, variable allocation is fundamentally and primarily a question of scope. If possible keep everything automatic and you don't have to do anything. If necessary, use dynamic allocation and take care to deallocate manually whenever appropriate.
(Edit: As #Oli says, you may also want to use dynamic allocation in a strictly local context at times, because most platforms limit the size of automatic variables to a much smaller limit than the size of dynamic memory. Think "huge array". Exceeding the available space for automatic variables usually has a colourful name such as "pile overrun" or something similar.)
In general, every call to malloc must have one corresponding call to free.* This has nothing to do with scope (i.e. nothing to do with functions or loops).
* Exceptions to this rule include using functions like strdup, but the principle is the same.
Broadly speaking, every pointer that is ever returned by malloc() must eventually be passed to free(). The scope of the variable that you store the pointer in does not affect this, because even after the variable is no longer in scope, the memory that the pointer points to will still be allocated until you call free() on it.
Well, the scope of the malloc'd memory lays between calls to malloc and free or otherwise until process is stopped (that is when OS cleans up for the process). If you never call free you get a memory leak. That could happen when address that you can pass to free goes out of scope before you actually used it - that is like loosing your keys for the car, car is still there but you can't really drive it. The error you are getting is most likely either because function returns a pointer to some memory that was not allocated using malloc or it returns a null pointer which you pass to free, which you cannot do.
You should free memory when you will no longer be accessing it. You should not free memory if you will be accessing it. This will give you a lot of pain.
If you don't want memory leak, you have to free the memory from malloc.
It can be very tricky. For example, if the // do some stuff has a continue, the free will be skipped and lead to memory leak. It is tricky, so we have shared_ptr in C++; and rumor has it salary of C programmer is higher than C++ programmer.
Sometimes we don't care memory leak. If the memory holds something that is needed during the whole lifetime of execution, you can choose not to free it. Example: a string for environment variable.
PS: Valgrind is a tool to help detect memory bugs. Especially useful for memory leak.
malloc(n) allocates n bytes of memory from a memory location named heap and then returns a void* type of pointer to it. The memory is allocated at runtime. Once you have allocated a memory dynamically, scope does not matter as long as you keep a pointer to it with you(or the address of it specifically). For example:
int* allocate_an_integer_array(int n)
{
int* p = (int*) (malloc(sizeof(int)*n));
return p;
}
This functions simply allocates memory from heap equal to n integers and returns a pointer to the first location. The pointer can be used in the calling function as you want to. The SCOPE does not matter as long as the pointer is with you..
free(p) returns the memory to heap.
The only thing you need to remember is to free it as if you don't free it and lose the value of its address, there will bw a memory leak. It is so because according to OS, you are still using the memory as you have not freed it and a memory leak will happen..
Also after freeing just set the value of the pointer to null so that u don't use it again as the same memory may be allocated again at any other time for a different purpose....
So, all you need to do is to be careful...
Hope it helps!
Related
I would be very glad if someone could help me in understanding completely the differences between the following codes:
// Code (1)
void f1 ( void ){
int * ptr1 = malloc ( sizeof(int) );
}
and
// Code (2)
void f2 ( void ){
int * ptr2 = malloc ( sizeof(int) );
free(ptr2);
}
As far I know, the instruction free is useful for deallocating the used memory, but on the other hand I know that every time we call a function g, if here there are new variables to be declared, they will be created and destroyed (i.e. deallocated, right?) after the execution of g.
Consequently:
do we need to use the instruction free in Code(2), or it is superfluous? (but maybe suggested for making the code more clear)
Thanks
Ps: you may be also interested in helping me with My previous related question. There, some users suggested to use the dynamic allocation of memory, and I am trying to understand why/how to.
No. malloc will not free the allocated memory. You need to use free to free the allocated chunk.
It is also the case that at the end of the program (main function) the allocated memory automatically freed by the system, but better to free it explicitly.
Always remember that the lifetime of dynamically allocated memory is till the end of the program if it is not deallocated specifically.
As said previously, memory allocated by malloc should be unallocated by free, or else you'll get memory leaks. What confuses you is that the ptr1var itself is unallocated (not the memory it's pointing to). The allocated memory is not a variable. So when you exit f1, you have no way to access your allocated memory any more, as you've lost the address.
BTW you've not explained why you need new addresses in your previous question.
No, you have to use free (or realloc)
From http://www.cplusplus.com/reference/cstdlib/malloc/:
If the function reuses the same unit of storage released by a deallocation function (such as free or realloc), the functions are synchronized in such a way that the deallocation happens entirely before the next allocation.
More information: How do malloc() and free() work?
No, malloc() will not release the memory when the function terminates.
If I may offer a suggestion though ...
If you want to allocate space (for a (very) small object) only throughout a (non-recursive) function you can use C99's variable length arrays
int foobar(size_t small_n) {
struct whatever autorelease[small_n]; // C99 VLA
// use autorelease
// no need to free
return 0;
} // memory for autorelease object released
In addition to other answers explaining about C dynamic memory allocation and the need to explicitly call free for each malloc-ed data (and when to call free is an issue; you want to avoid memory leaks), and you'll find a tool like valgrind helpful, you could also consider using Boehm's garbage collector. You basically could replace malloc by GC_malloc in your entire program and not bother a lot about explicit free-ing.
Of course, read the wikipage on garbage collection, and later the
GC handbook. It does have some caveats. So Boehm's GC is not a silver bullet.
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 am trying to undestand the C functions malloc and free. I know this has been discussed a lot on StackOverflow. However, I think I kind of know what these functions do by now. I want to know why to use them. Let's take a look at this piece of code:
int n = 10;
char* array;
array = (char*) malloc(n * sizeof(char));
// Check whether memory could be allocated or not...
// Do whatever with array...
free(array);
array = NULL;
I created a pointer of type char which I called array. Then I used malloc to find a chunk of memory that is currently not used and (10 * sizeof(char)) bytes large. That address I casted to type char pointer before assigning it to my previously created char pointer. Now I can work with my char array. When I am done, I'll use free to free that chunk of memory since it's not being used anymore.
I have one question: Why wouldn't I just do char array[10];? Wikipedia has only one small sentence to give to answer that, and that sentence I unfortunately don't understand:
However, the size of the array is fixed at compile time. If one wishes to allocate a similar array dynamically...
The slide from my university is similarily concise:
It is also possible to allocate memory from the heap.
What is the heap? I know a data structure called heap. :)
However, I've someone could explain to me in which case it makes sense to use malloc and free instead of the regular declaration of a variable, that'd be great. :)
C provides three different possible "storage durations" for objects:
Automatic - local storage that's specific to the invocation of the function it's in. There may be more than one instance of objects created with automatic storage, if a function is called recursively or from multiple threads. Or there may be no instances (if/when the function isn't being called).
Static - storage that exists, in exactly one instance, for the entire duration of the running program.
Allocated (dynamic) - created by malloc, and persists until free is called to free it or the program terminates. Allocated storage is the only type of storage with which you can create arbitrarily large or arbitrarily many objects which you can keep even when functions return. This is what malloc is useful for.
First of all there is no need to cast the malloc
array = malloc(n * sizeof(char));
I have one question: Why wouldn't I just do char array[10];?
What will you do if you don't know how many storage space do you want (Say, if you wanted to have an array of arbitrary size like a stack or linked list for example)?
In this case you have to rely on malloc (in C99 you can use Variable Length Arrays but for small memory size).
The function malloc is used to allocate a certain amount of memory during the execution of a program. The malloc function will request a block of memory from the heap. If the request is granted, the operating system will reserve the requested amount of memory.
When the amount of memory is not needed anymore, you must return it to the operating system by calling the function free.
In simple: you use an array when you know the number of elements the array will need to hold at compile time. you use malloc with pointers when you don't know how many elements the array will need to be at compile time.
For more detail read Heap Management With malloc() and free().
Imagine you want to allocate 1,000 arrays.
If you did not have malloc and free... but needed a declaration in your source for each array, then you'd have to make 1,000 declarations. You'd have to give them all names. (array1, array2, ... array1000).
The idea in general of dynamic memory management is to handle items when the quantity of items is not something you can know in advance at the time you are writing your program.
Regarding your question: Why wouldn't I just do char array[10];?. You can, and most of the time, that will be completely sufficient. However, what if you wanted to do something similar, but much much bigger? Or what if the size of your data needs to change during execution? These are a few of the situations that point to using dynamically allocated memory (calloc() or malloc()).
Understanding a little about how/when the stack and heap are used would be good: When you use malloc() or calloc(), it uses memory from the heap, where automatic/static variables are given memory on the stack, and are freed when you leave the scope of that variable, i.e the function or block it was declared in.
Using malloc and calloc become very useful when the size of the data you need is not known until run-time. When the size is determined, you can easily call one of these to allocate memory onto the heap, then when you are finished, free it with free()
Regarding What is the heap? There is a good discussion on that topic here (slightly different topic, but good discussion)
In response to However, I've someone could explain to me in which case it makes sense to use malloc() and free()...?
In short, If you know what your memory requirements are at build time (before run-time) for a particular variable(s), use static / automatic creation of variables (and corresponding memory usage). If you do not know what size is necessary until run-time, use malloc() or calloc() with a corresponding call to free() (for each use) to create memory. This is of course a rule-of-thumb, and a gross generalization. As you gain experience using memory, you will find scenarios where even when size information is known before run-time, you will choose to dynamically allocate due to some other criteria. (size comes to mind)
If you know in advance that you only require an array of 10 chars, you should just say char array[10]. malloc is useful if you don't know in advance how much storage you need. It is also useful if you need storage that is valid after the current function returns. If you declare array as char array[10], it will be allocated on the stack. This data will not be valid after your function returns. Storage that you obtain from malloc is valid until you call free on it.
Also, there is no need to cast the return value of malloc.
Why to use free after malloc can be understood in the way that it is a good style to free memory as soon as you don't need it. However if you dont free the memory then it would not harm much but only your run time cost will increase.
You may also choose to leave memory unfreed when you exit the program. malloc() uses the heap and the complete heap of a process is freed when the process exits. The only reason why people insist on freeing the memory is to avoid memory leaks.
From here:
Allocation Myth 4: Non-garbage-collected programs should always
deallocate all memory they allocate.
The Truth: Omitted deallocations in frequently executed code cause
growing leaks. They are rarely acceptable. but Programs that retain
most allocated memory until program exit often perform better without
any intervening deallocation. Malloc is much easier to implement if
there is no free.
In most cases, deallocating memory just before program exit is
pointless. The OS will reclaim it anyway. Free will touch and page in
the dead objects; the OS won't.
Consequence: Be careful with "leak detectors" that count allocations.
Some "leaks" are good!
Also the wiki has a good point in Heap base memory allocation:-
The heap method suffers from a few inherent flaws, stemming entirely
from fragmentation. Like any method of memory allocation, the heap
will become fragmented; that is, there will be sections of used and
unused memory in the allocated space on the heap. A good allocator
will attempt to find an unused area of already allocated memory to use
before resorting to expanding the heap. The major problem with this
method is that the heap has only two significant attributes: base, or
the beginning of the heap in virtual memory space; and length, or its
size. The heap requires enough system memory to fill its entire
length, and its base can never change. Thus, any large areas of unused
memory are wasted. The heap can get "stuck" in this position if a
small used segment exists at the end of the heap, which could waste
any magnitude of address space, from a few megabytes to a few hundred.
I am working on learning c. I understand that malloc() allocates a block of bytes that cannot be changed or corrupted without user request, however I find myself using it very often. To be exact, I am using malloc every time that I want to create either a struct or any of its contents that I want to reference in the future. I also do understand to free() the allocated memory when its complete.
Is my use of malloc correct?
Dynamic memory allocation (malloc and family) are there for two reasons:
Your data needs to persist beyond the scope that allocated it (e.g. multithreading)
Whatever you are allocating is too large for your stack
You should really be avoiding to allocate dynamic memory for any other reason. Automatic (stack) variables are far less prone to errors and are automatically deallocated for you at the end of the scope.
Having "corrupted memory" like you call it can only really arise from bad programming and can happen on both the stack and the heap and you should not rely on dynamic memory to provide safety from buffer overflows or other mistakes that lead to memory corruption.
There is a reason why many functions in the C standard library get a pointer to a buffer as an argument to put results in: it allows you to allocate those buffers on your stack. e.g:
ssize_t read(int fd, void *buf, size_t count);
Also as mentioned by another answer: Your stack memory is already in the CPU cache and is thus far faster accessible.
Please also consider the other types of allocation:
int foo;
outside of a block will allocate a global variable, which is alive during the whole lifetime of your process, and visible for other modules of the program.
static int foo;
outside of a block is the same but visible in the actual module only.
int foo;
inside a block is alive only while the code in the block runs, then it's destroyed.
static int foo;
inside a block is visible in the block only, but it preserves its value for the entire lifetime of the process.
I'm doing a lot of embedded C coding, and using malloc() is absolutely prohibited. And it's entirely possible. You typically need malloc() if you don't know the size of your problem at compile time. But even in some cases like that, you can replace dynamic memory allocation with other techinques like recursion, line-based processing etc, etc.
It depends on what you mean by
cannot be changed or corrupted without user request
If you are referring to code - then it's usually called client, not user. And it's still unclear what do you mean by that. But that's not the point.
The point is that malloc() is one of the functions used for dynamic memory allocation. It means that you can pass an address returned by this function somewhere else and data stored there will be there until it's manually deallocated. Unlike static memory allocation which is automatically freed when it's out of the scope.
So, you probably shouldn't be using malloc() if memory allocated by it is freed in the same scope, just because it's meaningless and because static allocation is faster because it's easier for CPU to cache and it's initialized at program startup, not at runtime as heap allocated memory.
I am mainly a Java person recently working on some projects involving C so please bear with me if it's a basic C question.
So inside my main I have a while loop and I declare a variable each iteration.
int main()
{
int done = 0;
while(!done)
{
char input[1024];
scanf("%s", input);
//parse the input string
...
}
}
Now since the input variable will change every time depending on what the user wants I have to use a "new" variable each time. However, I think the above declaration will be causing memory leak ultimately(or will it?). I would like to know if gcc takes care of garbage collection.
Is there any better approach without allocating and freeing after every iteration?
I think the above declaration will be causing memory leak ultimately(or will it?).
No, it wouldn't: input is an automatic (AKA "Stack") variable, it will get "deallocated" as soon as it goes out of scope (i.e. after the closing brace).
Is there any better approach without allocating and freeing after every iteration?
There is no actual allocation or deallocation going on: the space in the automatic memory (AKA "on the stack") is allocated by some compile-time bookkeeping around the stack pointer. The access of automatic variables is a very fast operation heavily assisted by hardware, so there is no loss of efficiency there.
Dynamic memory allocation (Java-style) is done with malloc/calloc/realloc in C. These are not garbage collected - you need to explicitly free every pointer that you allocated.
However, I think the above declaration will be causing memory leak ultimately(or will it?)
It will not. The inputobject is discarded every time the trailing } of the loop is encountered.
As other have already said, stack variables are automatically freed when they go out of scope and are usually overwritten soon.
That being said, the reason I'm writing this answer is to stress this: there is no garbage collector in C, at least not by default. This means that heap allocated memory (usually initialized with malloc, calloc and others) must be manually freed by you (using free).