In my project have a sub-function. In this function, I need to store my data temporarily. So I use malloc(), I'm not sure whether is necessary to use free()?
void *hash2(unsigned char *mes, element_t d)
{
size_t iterations = strlen(mes) / 8;
unsigned char *rtemp = malloc(32 * sizeof(char));
SHA256(mes, iterations, rtemp);
element_from_hash(d, rtemp, 32);
free(rtemp);
}
As already stated in the present answer, you should free any memory that is no longer needed, if you allocate memory within the function, not freeing it and not returning any pointer to it will cause a memory leak.
Note that your function *hash2(...), having void* return type, must return a value, if you don't need it to, use void instead.
In your particular code it does seem that you wouldn't need to use malloc anyway, you can use a local array unsigned char rtemp[32];. malloc is a heavy function that involves system calls, if you can avoid it, you should.
Yes, as you are not using the allocated memory anymore, you must free it.
Once you return from the function call, you'll have no way to access the allocated memory, hence it cannot be freed up. So, you need to to pass the pointer to free() before leaving the function scope (as seen in the snippet) to avoid memory leak.
Related
So i want to return an array of a size n (variable) which my function has as input. I know that in order to return arrays in C I have to define them static, but the problem is that n is a variable and thus I get an error. I thought of actually using malloc/calloc but then I won't be able to free them after the function returns the array. Please take note that I'm not allowed to change anything on main(). Are there any other alternatives which I could use? Thanks in advance.
float *Arr( int *a , int n ){
static float b[ n ];
return b
}
Got to point out that the function will only be called Once,I saw the solution you posted but i noticed you aren't freeing the allocated memory,is it not of much importance when the malloc is called inside a function?
The important thing to notice here is that this syntax:
float arr[n];
Allocates an array on the stack of the current function. In other words, that array is a local variable. Any local variable becomes invalid after the function returns, and therefore returning the array directly is undefined behavior. It will most likely cause a crash when trying to access the array from outside the function, if not anything worse.
In addition to that, declaring a variable-length array as static is invalid in any case.
If you want to write a function which creates and returns any kind of array (dynamically sized or not), the only option you have is to use dynamic allocation through malloc() and then return a pointer to the array (technically there's also alloca() to make dynamic stack allocations, but I would avoid it as it can easily break your program if the allocation is too large).
Here's an example of correct code:
float *create_array(size_t n_elements){
float *arr = malloc(sizeof(float) * n_elements);
if (arr == NULL) {
// Memory could not be allocated, handle the error appropriately.
}
return arr;
}
In this case, malloc() is reserving memory outside of the local stack of the function, in the heap. The result is a pointer that can be freely returned and passed around without any problem, since that area of memory keeps being valid after the function returns (until it is released). When you're done working with the data, you can release the allocated memory by calling free():
float *arr = create_array(100);
// ...
free(arr);
If you don't have a way to release the memory through free() after using malloc(), that's a problem in the long run, but in general, it is not a strict requirement: if your array is always needed, from its creation until the exit of the program, then there's no need to explicitly free() it, since memory is automatically released when the program terminates.
If your function needs to be called more than once or needs to create significantly sized arrays that are only useful in part of the program and should therefore be discarded when no longer in use, then I'm afraid there's no good way of doing it. You should use free() in that case.
To answer your question precisely:
Please take note that I'm not allowed to change anything on main(). Are there any other alternatives which I could use?
No, there are no other better alternatives. The only correct approach here is to dynamically allocate the array through malloc(). The fact that you cannot free it afterwards is a different kind of problem.
So i want to return an array of a size n(variable) which my function
has as input,
You can't, because C functions cannot return arrays at all. They can, and some do, return pointers, however, as your function is declared to do. Such a pointer may point to an element of an array.
i know that in order to return arrays in c i have to
define them static,
As long as I am being pedantic, the problem is to do with the lifetime of the object to which the returned pointer points. If it is an element of an automatically-allocated array, then it, along with the rest of the array, ceases to exist when the function returns. The caller must not try to dereference such a pointer.
The two other alternatives are
static allocation, which you get by declaring the variable static or by declaring it at file scope, and
dynamic allocation, which you get by reserving memory via malloc(), calloc(), or a related function.
Statically allocated objects exist for the entire lifetime of the program, and dynamically allocated ones exist until deallocated.
but problem is that n is a variable and thus i get
an error.
Yes, because variable-length arrays must be automatically allocated. Static objects exist for the whole run of the program, so the compiler needs to reserve space for them at compile time.
I thought of actually using malloc/calloc but then i won't be
able to free them after the function returns the array.
That's correct, but dynamic allocation is still probably the best solution. It is not unreasonable for a called function to return a pointer to an allocated object, thus putting the responsibility on its caller to free that object. Ordinarily, that would be a well-documented characteristic of the function, so that its callers know that such responsibility comes with calling the function.
Moreover, although it's a bit untidy, if your function is to be called only once then it may be acceptable to just allow the program to terminate without freeing the array. The host operating system can generally be relied upon to clean up the mess.
Please take
note that im not allowed to change anything on main(),are there any
other alternatives which i could use?
If you have or can impose a bound on the maximum value of n then you can declare a static array of that maximum size or longer, and return a pointer to that. The caller is receiving a pointer, remember, not an array, so it can't tell how long the pointed-to array actually is. All it knows is that the function promises n accessible elements.
Note well that there is a crucial difference between the dynamic allocation and static allocation alternatives: in the latter case, the function returns a pointer to the same array on every call. This is not inherently wrong, but it can be problematic. If implemented, it is a characteristic of the function that should be both intentional and well-documented.
If want an array of n floats where n is dynamic, you can either create a
variadic-length array (VLA):
void some_function(...)
{
//...
float b[ n ]; //allocate b on the stack
//...
}
in which case there would be no function call for the allocation, or you can allocate it dynamically, e.g., with malloc or calloc, and then free it after you're done with it.
float *b = malloc(sizeof(*b)*n);
A dynamic (malloc/calloc) allocation may be wrapped in a function that returns a pointer to the allocated memory (the wrapper may do some initializations on the allocated memory after the memory has been successfully allocated). A VLA allocation may not, because a VLA ends its lifetime at the end of its nearest enclosing block (C11 Standard - 6.2.4 Storage durations of objects(p7)).
If you do end up wrapping a malloc/calloc call in a "constructor" function like your float *Arr(void), then you obviously should not free the to-be-returned allocated memory inside Arr–Arr's caller would be responsible for freeing the result (unless it passed the responsibility over to some other part of the program):
float *Arr( int n, ...
/*some params to maybe initialize the array with ?*/ )
{
float *r; if (!(r=malloc(sizeof(*r)*n)) return NULL;
//...
//do some initializations on r
//...
return r; //the caller should free it
}
you could use malloc to reserve memory for your n sized array
Like this:
#include <stdlib.h>
#include <stdio.h>
float * arr(int * a, int n ) {
float *fp = malloc ( (size_t) sizeof(float)*n);
if (!fp) printf("Oh no! Run out of memory\n");
return fp;
}
int main () {
int i;
float * fpp = arr(&i,200);
printf("the float array is located at %p in memory\n", fpp);
return(0);
}
It seems like what you want to do is:
have a function that provides (space for) an array with a variable number of elements,
that the caller is not responsible for freeing,
that there only needs to be one instance of at a time.
In this case, instead of attempting to define a static array, you can use a static pointer to manage memory allocated and freed with realloc as needed to adjust the size, as shown in the code below. This will leave one instance in existence at all times after the first call, but so would a static array.
This might not be a good design (it depends on circumstances not stated in the question), but it seems to match what was requested.
#include <stdio.h>
#include <stdlib.h>
float *Arr(int *a , int n)
{
// Keep a static pointer to memory, with no memory allocated initially.
static float *b = NULL;
/* When we want n elements, use realloc to release the old memory, if any,
and allocate new memory.
*/
float *t = realloc(b, n * sizeof *t);
// Fail if the memory allocation failed.
if (!t)
{
fprintf(stderr, "Error, failed to allocate memory in Arr.\n");
exit(EXIT_FAILURE);
}
// Return the new memory.
return b;
}
I am learning the basic of C programming and find it quite strange and difficult. Especially with dynamic memory allocation, pointer and similar things. I came upon this function and don't quite understand what is wrong with it.
char *strdup(const char *p)
{
char *q;
strcpy(q, p);
return q;
}
I think I have to malloc and free q. But the function " return q". Doesn't that mean that it will store q value in its own memory. So the data still be saved after the function executed?
When it is appropriate to use malloc? As I understand so far is that I have to malloc a new variable every time I need that variable declared in a function to be used elsewhere. Is that true? Is there any other situation where malloc is needed?
The type of q is a pointer, and pointers hold addresses -- so what you are returning is the address that pointer holds.
Until you give that pointer a valid address, it points off to who-knows-where, memory that you may or may not own and have the right to access. So, the strdup call will copy a string from the address held in p into some location you probably don't own.
If you had done a malloc first, and given q the results of the malloc, then q would hold a valid address, and your strdup would put the copy into memory that you did own (assuming you malloc'd enough space for the string -- a strlen on p would tell you how much you needed).
Then, when you returned q, you would be giving the caller the address as well. Any code with that address can see the string you put there. If some future code were to free that address, then what it holds is up in the air -- it could be anything at all.
So, you don't want to free q before you return the address that it holds -- you need to let the caller free the address it gets from you, when it is ready to do so.
In terms of when you malloc, yes, if you want to return an address that will remain viable after your function completes, you need to malloc it -- giving the caller the address of a local variable, for example, would be bad: the memory is freed when the function returns, you don't own it anymore.
Another typical use of malloc is for building up dynamic data structures like trees and lists -- you can't know how much memory you need up front, so you build the list or tree up as you need to, malloc'ing more memory for each node in the structure.
My personal rules are: use malloc() when an object is too big to put on the stack or/and when it has to live outside the scope of the current block. In your case, I believe, you should do something like the following:
char *strdup(const char *p)
{
char *q;
q = malloc(strlen(p) + 1);
if(NULL != q)
strcpy(q, p);
return q;
}
malloc(X) creates space (of size X bytes) on the heap for you to play with. The data that you write to these X byes stays put when your function returns, as a result, you can read what your strdup() function wrote to that space on the heap.
free() is used for freeing space on the heap. You can pass a pointer to this function that you obtained only as a result of a malloc() or realloc() call. This function may not clear out the data, it just means that a subsequent call to malloc() may return the address of the same space that you just freed. I say "may" because these are all implementaion defined and should not be relied upon.
In the piece of code you wrote, the strcpy() function copies the bytes one by one from p to q until it finds a \0 at the location pointed to by q (and then it copies the \0 as well). In order to write data somewhere, you need to allocate space first for the data to be written, hence, one option is that you call malloc() to create some space and then write data there.
Well, calling free() is not mandatory as your OS will reclaim the space allocated by malloc() when you program ends, but as long as your program runs, you may be ocupying more space than you need to - and that's bad for your program, for other programs, for the OS and the universe as a whole.
I think I have to malloc and free q. But the function " return q". Doesn't that mean that it will store q value in its own memory. So the data still be saved after the function executed?
No, your data won't be saved. In fact your pointer q is being used without allocating it's size can cause problems. Also, once this function execution complete, variable char* q will be destroyed.
You need to allocate memory to pointer q before copying data as suggested by #Michael's answer.But once you finish using the data return by this function, you will need to manually free() the memory you allocated or else it will cause memory leak (a situation where the memory is allocated but there is no pointer refer to that chunk of memory you allocated and hence will be inaccessible throughout program execution)
char *strdup(const char *p) // From #Michael's answer
{
char *q;
q = malloc(strlen(p) + 1);
if(NULL != q)
strcpy(q, p);
return q;
}
void someFunction()
{
char* aDupString = strdup("Hello World!!!");
/*... somecode use aDupString */
free(aDupString); // If not freed, will cause memory leaks
}
When it is appropriate to use malloc? Is there any other situation where malloc is needed?
It is appropriate to use in following situations:
1> The usage size of array are unknown at compile time.
2> You need size flexibility. For example, your function need to work with small data size and large data size. (like Data structure such as Link list,Stacks, Queues, etc.)
As I understand so far is that I have to malloc a new variable every time I need that variable declared in a function to be used elsewhere. Is that true?
I think this one is partially true. depending on what you are trying to achive, there might be a way to get around using malloc though. For example, your strdup can also be rewrite in following way:
void strdup2(const char *p, char* strOut)
{
// malloc not require
strcpy(strOut, p);
}
void someFunction()
{
char aString[15] = "Hello World!!!";
char aDupStr[sizeof(aString)];
strdup2(aString, aDupStr);
// free() memory not required. but size is not dynamic.
}
I am trying to return an array using malloc in a function:
char* queueBulkDequeue(queueADT queue, unsigned int size)
{
unsigned int i;
char* pElements=(char*)malloc(size * sizeof(char));
for (i=0; i<size; i++)
{
*(pElements+i) = queueDequeue(queue);
}
return pElements;
}
The problem is that I need to free it because my MCU's heap size is limited. But I want to return it so I cannot free it in the function, right?. Can I free the allocated memory outside the function (where I call the function). Is there any best practices for this? Thank you in advance!
As the memory allocated by malloc() is on the heap and not on the stack, you can access it regardless of which function you are in. If you want to pass around malloc()'ed memory, you have no other option than freeing it from the caller. (in reference counting terms, that's what is called an ownership transfer.)
1) Yes, you can free() the malloc'ed memory outside the function
2) No, you cannot free it inside the function and have the data passed outside the function, so you must do 1) here
3) If you're concerned about scarce memory, you need to check for failure from memory allocations always, which you fail to do here, which is then likely to lead to a segfault
Ofcourse you can free the memory allocated in a function outside of that function provided you return it.
But, an alternative would be to modify your function like below, where the caller only allocates & frees the memory. This will be inline with concept of the function which allocates the memory takes responsibility for freeing the memory.
void queueBulkDequeue(queueADT queue, char *pElements, unsigned int size)
{
unsigned int i;
for (i=0; i<size; i++)
{
*(pElements+i) = queueDequeue(queue);
}
return;
}
//In the caller
char *pElements = malloc(size * sizeof(char));
queueBulkDequeue(queue, pElements, size);
//Use pElements
free(pElements);
Yes, you can free memory allocated in a function that you call outside the function; this is precisely what you need to do in this case.
Alternatives include passing a buffer and its length into the function, and returning the actual length to the caller, the way fgets does. This may not be the best alternative, because the callers would need to call your function in a loop.
I've read conflicting information on the internet about this. To the best of my knowledge all variables in a function only exist for the life time of the function and so this shouldn't be necessary. But with dynamic pointer arrays I'm not sure.
Thanks
You don't free pointers - they are cleaned up automatically. What you need to free is the memory that was acquired by malloc. You use pointers to access that memory.
Put laconically: Every pointer you get from malloc has to go to free exactly once. Nothing more, nothing less.
It depends on where the memory referenced by the pointer has been allocated.
There are fundamentally 2 ways of allocating space in C.
Using the stack :
void foo(){
int stack_variable = 10;
int *stack_pointer = &stack_variable; //bar shouldn't be freed.
}
Using the heap:
void foo(){
int * heap_pointer = (int *) malloc(sizeof(int)); //heap_pointer need to be freed somewhere
}
In the first case there is no problem: memory allocated in the stack will be released when the function returns. Even if you are using a pointer, and it points to some data in the stack, free isn't necessary. Be careful to not use a pointer to data allocated in the stack when the data itself goes out of scope:
int * foo(){
int stack_variable = 10;
int *ptr = &stack_variable;
return ptr;
}
int * ptr = foo(); // you have a problem here: ptr points to a memory region that no longer exist!
In the second case you are using the heap so you need to use free somewhere, to explicitly release it.
If you have allocated data using malloc/calloc, you need to free the data.
Addition due to curious comment by #Julia Childe:
Well, the point of allocating dynamic memory is that it will remain there till you explicitly free it. This enables you to pass pointers both from and to functions and you are not limited by the scope of a specific function, i.e. main.
Thereby, you can allocate memory for data when you need to and not in advance, thus dynamic memory.
If we did not have this ability, we would have to know how much memory space we would use at compile time.
Hope this clears out some question marks.
I am working with my first straight C project, and it has been a while since I worked on C++ for that matter. So the whole memory management is a bit fuzzy.
I have a function that I created that will validate some input. In the simple sample below, it just ignores spaces:
int validate_input(const char *input_line, char** out_value){
int ret_val = 0; /*false*/
int length = strlen(input_line);
out_value =(char*) malloc(sizeof(char) * length + 1);
if (0 != length){
int number_found = 0;
for (int x = 0; x < length; x++){
if (input_line[x] != ' '){ /*ignore space*/
/*get the character*/
out_value[number_found] = input_line[x];
number_found++; /*increment counter*/
}
}
out_value[number_found + 1] = '\0';
ret_val = 1;
}
return ret_val;
}
Instead of allocating memory inside the function for out_value, should I do it before I call the function and always expect the caller to allocate memory before passing into the function? As a rule of thumb, should any memory allocated inside of a function be always freed before the function returns?
I follow two very simple rules which make my life easier.
1/ Allocate memory when you need it, as soon as you know what you need. This will allow you to capture out-of-memory errors before doing too much work.
2/ Every allocated block of memory has a responsibility property. It should be clear when responsibility passes through function interfaces, at which point responsibility for freeing that memory passes with the memory. This will guarantee that someone has a clearly specified requirement to free that memory.
In your particular case, you need to pass in a double char pointer if you want the value given back to the caller:
int validate_input (const char *input_line, char **out_value_ptr) {
: :
*out_value_ptr =(char*) malloc(length + 1); // sizeof(char) is always 1
: :
(*out_value_ptr)[number_found] = input_line[x];
: :
As long as you clearly state what's expected by the function, you could either allocate the memory in the caller or the function itself. I would prefer outside of the function since you know the size required.
But keep in mind you can allow for both options. In other words, if the function is passed a char** that points to NULL, have it allocate the memory. Otherwise it can assume the caller has done so:
if (*out_value_ptr == NULL)
*out_value_ptr =(char*) malloc(length + 1);
You should free that memory before the function returns in your above example. As a rule of thumb you free/delete allocated memory before the scope that the variable was defined in ends. In your case the scope is your function so you need to free it before your function ends. Failure to do this will result in leaked memory.
As for your other question I think it should be allocated going in to the function since we want to be able to use it outside of the function. You allocate some memory, you call your function, and then you free your memory. If you try and mix it up where allocation is done in the function, and freeing is done outside it gets confusing.
The idea of whether the function/module/object that allocates memory should free it is somewhat of a design decision. In your example, I (personal opinion here) think it is valid for the function to allocate it and leave it up to the caller to free. It makes it more usable.
If you do this, you need to declare the output parameter differently (either as a reference in C++ style or as char** in C style. As defined, the pointer will exist only locally and will be leaked.
A typical practice is to allocate memory outside for out_value and pass in the length of the block in octets to the function with the pointer. This allows the user to decide how they want to allocate that memory.
One example of this pattern is the recv function used in sockets:
ssize_t recv(int socket, void *buffer, size_t length, int flags);
Here are some guidelines for allocating memory:
Allocate only if necessary.
Huge objects should be dynamically
allocated. Most implementations
don't have enough local storage
(stack, global / program memory).
Set up ownership rules for the
allocated object. Owner should be
responsible for deleting.
Guidelines for deallocating memory:
Delete if allocated, don't delete
objects or variables that were not
dynamically allocated.
Delete when not in use any more.
See your object ownership rules.
Delete before program exits.
In this example you should be neither freeing or allocating memory for out_value. It is typed as a char*. Hence you cannot "return" the new memory to the caller of the function. In order to do that you need to take in a char**
In this particular scenario the buffer length is unknown before the caller makes the call. Additionally making the same call twice will produce different values since you are processing user input. So you can't take the approach of call once get the length and call the second time with the allocated buffer. Hence the best approach is for the function to allocate the memory and pass the responsibility of freeing onto the caller.
First, this code example you give is not ANSI C. It looks more like C++. There is not "<<" operator in C that works as an output stream to something called "cout."
The next issue is that if you do not free() within this function, you will leak memory. You passed in a char * but once you assign that value to the return value of malloc() (avoid casting the return value of malloc() in the C programming language) the variable no longer points to whatever memory address you passed in to the function. If you want to achieve that functionality, pass a pointer to a char pointer char **, you can think of this as passing the pointer by reference in C++ (if you want to use that sort of language in C, which I wouldn't).
Next, as to whether you should allocate/free before or after a function call depends on the role of the function. You might have a function whose job it is to allocate and initialize some data and then return it to the caller, in which case it should malloc() and the caller should free(). However, if you are just doing some processing with a couple of buffers like, you may tend to prefer the caller to allocate and deallocate. But for your case, since your "validate_input" function looks to be doing nothing more than copying a string without the space, you could just malloc() in the function and leave it to the caller. Although, since in this function, you simply allocate the same size as the whole input string, it almost seems as if you might as well have the caller to all of it. It all really depends on your usage.
Just make sure you do not lose pointers as you are doing in this example
Some rough guidelines to consider:
Prefer letting the caller allocate the memory. This lets it control how/where that memory is allocated. Calling malloc() directly in your code means your function is dictating a memory policy.
If there's no way to tell how much memory may be needed in advance, your function may need to handle the allocation.
In cases where your function does need to allocate, consider letting the caller pass in an allocator callback that it uses instead of calling malloc directly. This lets your function allocate when it needs and as much as it needs, but lets the caller control how and where that memory is allocated.