1)
For which datatypes must I allocate memory with malloc?
For types like structs, pointers, except basic datatypes, like int
For all types?
2)
Why can I run this code? Why does it not crash? I assumed that I need to allocate memory for the struct first.
#include <stdio.h>
#include <stdlib.h>
typedef unsigned int uint32;
typedef struct
{
int a;
uint32* b;
}
foo;
int main(int argc, char* argv[])
{
foo foo2;
foo2.a = 3;
foo2.b = (uint32*)malloc(sizeof(uint32));
*foo2.b = 123;
}
Wouldn't it be better to use
foo* foo2 = malloc(sizeof(foo));
3)
How is foo.b set? Does is reference random memory or NULL?
#include <stdio.h>
#include <stdlib.h>
typedef unsigned int uint32;
typedef struct
{
int a;
uint32* b;
}
foo;
int main(int argc, char* argv[])
{
foo foo2;
foo2.a = 3;
}
All types in C can be allocated either dynamically, automatically (on the stack) or statically. The issue is not the type, but the lifetime you want - you use malloc when you want an object to exist outside of the scope of the function that created it, or when you don't know in advance how big a thing you need.
Edit to address your numbered questions.
There are no data types you must allocate with malloc. Only if you want a pointer type to point to valid memory must you use the unary & (address-of) operator or malloc() or some related function.
There is nothing wrong with your code - the line:
foo foo2;
Allocates a structure on the stack - then everything works as normal. Structures are no different in this sense than any other variable. It's not better or worse to use automatic variables (stack allocation) or globals or malloc(), they're all different, with different semantics and different reasons to choose them.
In your example in #3, foo2.b's value is undefined. Any automatic variable has an undefined and indeterminate value until you explicitly initialize it.
You must allocate with malloc any memory that you wish to be managed manually, as opposed to automatically. It doesn't matter if what's stored there is an int or a double or a struct or anything; malloc is all about manual memory management.
When you create a variable without malloc, it is stored on the stack and when it falls out of scope, its memory is automatically reclaimed. A variable falls out of scope when the variable is no longer accessible; e.g. when the block or function that the variable was declared in ends.
When you allocate memory with malloc, it is stored on the heap, and malloc returns a pointer to this memory. This memory will not be reclaimed until you call free on it, regardless of whether or not a pointer to it remains accessible (when no pointers remain to heap-allocated memory, this is a memory leak). This means that you gain the ability to continue to use the memory you allocated after the block or function that it was allocated in ends, but on the other hand you now have the responsibility to manually deallocate it when you are finished with it.
In your example, foo2 is on the stack, and will be automatically deallocated when main ends. However, the memory pointed to by foo2.b will not be automatically deallocated, since it is on the heap. This isn't a problem in your example because all memory is returned to the OS when the program ends, but if it were in a function other than main it would have been a memory leak.
foo foo2; automatically allocates the structure on the stack, and it is automatically deallocated when the enclosing function (main in this case) ends.
You only need to allocate memory on the heap, using malloc, if you need the structure to persist after the enclosing scope ends. You may also need to do this when the object is too large to fit on the stack.
2) Why can I run this code? Why does it not crash?
The code never refers to undefined memory or NULL. Why would it crash? (You have a memory leak as written, but it's probably because you're only showing part of the code and in the given program it's not a problem anyway.)
The alternative code you suggest will also work, though memory returned from malloc is also uninitialized by default. (I once worked with a custom memory allocator that filled returned memory blocks with ? characters by default. Still perfectly legal by the rules. Note that calloc returns a pointer to zero-initialized memory; use it if that's what you want.)
3) How is foo.b set? Does is reference random memory or NULL?
Random memory. Stack allocated structures are not initialized for you.
You can do it, but this is not enough.
Because, the second field is a pointer, which has to be set to a valid address. One of the ways to do it is by allocating memory (with malloc).
To your first question -
use malloc when you MUST manage object's lifetime manually.
For instance, the second code could be rewritten as:
int main(int argc, char* argv[])
{
foo foo2; uint32 b;
foo2.a = 3;
foo2.b = &b;
*foo2.b = 123;
}
This is better, because lifetime is the same, and the memory is now on stack - and doesn't need to be freed.
The memory for the struct instance ("foo2") will be allocated on the stack - there's no need to allocate memory for this yourself - if you do allocate using malloc be sure to free off the memory at a later date.
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;
}
This question already has answers here:
What and where are the stack and heap?
(31 answers)
Closed 7 years ago.
In C, is there a difference between
struct Foo foo;
struct Foo* fooPtr = &foo;
and
struct Foo* fooPtr = (struct Foo*) malloc(sizeof(struct Foo));
Of course there is.
It depends on where you allocate the foo variable - it may be in the (statically allocated) data segment or just on the stack. In the first case, it will never go away, and in the second case, it might go away too soon, depending on what you do with the pointer.
OTOH, if you malloc(), the data is put onto the heap, and it will stay until you free() it. This allows you to be flexible with your memory.
Yes There is a difference.
Because of dynamic memory allocation in the second case, it will be allocated in heap section of memory and you need to free it explicitly, How ever in first case you do not need to take care of that as it is allocated in the stack section of Memory.
malloc-ed memory must be free-ed, stack and global memory doesn't need to be. malloc-ed memory lives until free-ed, stack memory goes away when the function returns. malloc-ed memory can be dynamically sized at runtime, global memory cannot (and while in C99 and higher, stack memory can be dynamically sized, there are limits as far as stability and performance goes).
Sure there is. It has to do with how long the memory for that variable is alive.
int * bad_code() {
int x;
return &x;
}
int good_code() {
int *x = malloc(sizeof(int));
return x;
}
Using the memory referred to by bad_code() is not allowed. But you can continue to use the memory referred to by good_code().
Yes the first is stored either on the stack if declared as a function local or in data memory if declared as a file global. So that your pointer is pointing to this part of memory.
In the latter case, malloc allocates memory on the heap and the pointer points to memory on the heap. This must be freed as stated by others.
What is the difference between declaring an array "dynamically",
[ie. using realloc() or malloc(), etc... ]
vs
declaring an array within main() with Global scope?,
eg.
int main()
{
int array[10];
return 0;
}
I am learning, and at the moment it feels that there is not much differnce between
declaring a variable (array, whatever) -with Global scope,
when compared to a
dynamically allocated variable (array, whatever) -AND never calling free() on it AND allowing it to be 'destoryed' when the program ends'
What are the consequences of either option?
EDIT
Thank you for your responses.
Global scope should have been 'local scope' -local to main()
When you declare an array like int arr[10] in a function, the space for the array is allocated on the stack. The memory will be freed when your function exits.
When you declare an array or any other data structure using malloc() or realloc(), you allocated the space on the heap and the memory will only be freed afer the program exits. So when the program is running, you are responsible for freeing it using free() after you no longer want to use it. If you don't free it and make your array pointer point to something else, you will create a memory leak. However, your computer will always be able to retrieve all the program's used memory after the program ends because of virtual memory.
As kaylum said in comment below your question, the array in your second example does not have global scope. Its scope is limited to main(), and it is inaccessible in other scopes unless main() explicitly makes it available (e.g. passes it by argument to another function).
Dynamic memory allocation means that the programmer explicitly allocates memory when needed, and explicitly releases it when no longer needed. Because of that, the amount of memory allocated can be determined at run time (e.g. calculated from user input). Also, if the programmer forgets to release the memory, or reallocates it inappropriately, memory can be leaked (still allocated by the program, but not accessible by the program). For example;
/* within a function */
char *p = malloc(100);
p = malloc(200);
free(p);
leaks 100 bytes, every time this code is executed, because the result of the first malloc() call is never released, and it is then inaccessible to the program because its value is not stored anywhere.
Your second example is actually an array of automatic storage duration. As far as your program is concerned, it only exists until the end of the scope in which it is created. In your case, as main() returns, the array will cease to exist.
An example of an array with global scope is
int array[10];
void f() {array[0] = 42;}
int main()
{
array[0] = 10;
f();
/* array[0] will be 42 here */
}
The difference is that this array exists and is accessible to every function that has visibility of the declaration, within the same compilation unit.
One other important difference is that global arrays are (usually) zero initialised - a global array of int will have all elements zero. A dynamically allocated array will not have elements initialised (unless created with calloc(), which does initialise to zero). Similarly, an automatic array will not have elements initialised. It is undefined behaviour to access the value of something (including an array element) that is uninitialised.
So
#include <stdio.h>
int array[10];
int main()
{
int *array2;
int array3[10];
array2 = malloc(10*sizeof(*array2));
printf("%d\n", array[0]); /* okay - will print 0 */
printf("%d\n", array2[0]); /* undefined behaviour. array2[0] is uninitialised */
printf("%d\n", array3[0]); /* undefined behaviour. array3[0] uninitialised */
return 0;
}
Obviously the way to avoid undefined behaviour is to initialise array elements to something valid before trying to access their value (e.g. printing them out, in the example above).
Is it allowed to dynamically allocate memory for static variable like this:
#include <stdio.h>
#include <stdlib.h>
struct person
{
int age;
int number;
};
static struct person* person_p = NULL;
int main()
{
person_p = (struct person*)malloc(10 * sizeof(struct person));
}
The above code built, but is it really allowed to dynamically allocate memory for static variable?
Yes, it's valid and allowed. (Unless you're using the pointer as a placeholder) You can (and need to) dynamically allocate and free() memory to and from the pointer before and after using it.
Rather, please make a note, you do not cast the return value of malloc() and family in C.
I don't see why not. Even though static means there can only be one instance of the object, you still need space for that object. Keep in mind however that anything that is malloc'd needs to be free'd, so you will want to do that at the end of your main() function.
Memory isn't "owned" by pointers to it. You can do the following things:
Dynamically allocate memory
Make a pointer point to that memory
It doesn't really make sense to say "dynamically allocate memory for a pointer".
Any pointer can point to any object (subject to alignment and aliasing restrictions), it makes no difference what the storage duration of the pointer is.
Note that it is the pointer that is static, not the memory it points to.
static means two unrelated things:
give static memory allocation (memory allocated at start of program, and only released at end of program)
give internal linkage (do not allow other compilation units - modules - to access the identifier. Here a variable, could be a function)
In your code, 1. static memory allocation is irrelevant, since the variable is global anyway and as such already has it.
Then, 2. internal linkage does not matter either, because what you are trying to do is inside the module anyway.
In other words, person_p is exactly as a usual global variable within your module, and you can do whatever you want to it.
It's only the pointer that is defined by this line of code, so you can dynamically allocate memory elsewhere, and assign the memory address to person_p if you wish.
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.