What happens with the memory when decrease array with realloc? - c

I was wondering what happens with the memory when u realloc -1 your array. According everything that I've read about realloc I suppose that pointer still points at the same place in memory (there's no need for function to seek another block of memory as that one is available and sufficient), tell me if I'm wrong. My question is: Is the deleted piece of array deleted (like with using free()) or are the values stay untouched and the piece of memory is being shared for future operations of malloc, calloc etc.?
EDIT:
I have one more question. Does this function work properly? It should delete element of array previously overwriting it by the next element of the array. Doing it over the whole array, the last element is the same as the one before last and the last one is deleted. PicCounter is the number of pictures already uploaded to program. Check this out:
int DeletePicture(struct Picture **tab, int *PicCounter)
{
int PicToDelete;
printf("Enter the number of pic to delete ");
scanf("%d", &PicToDelete);
for (int i = PicToDelete - 1; i < (*PicCounter) - 2; i++)
{
(*tab)[i] = (*tab)[i + 1];
}
struct Picture *temp;
temp = realloc(*tab, ((*PicCounter)-1) * sizeof(*temp));
if (temp != NULL)
{
*tab = temp;
//That doesn't delete the element, because in main I can still print it
//like e.g. tab[lastelement].
(*PicCounter)--;
printf("Picture has been deleted\n");
return 0;
}
else
{
printf("Memory reallocation error\n");
return 1;
}
}

Regarding void *realloc(void *ptr, size_t size), the C standard says in C 2018 7.22.3.5 paragraphs 2 and 3:
The realloc function deallocates the old object pointed to by ptr and returns a pointer to a new object that has the size specified by size. The contents of the new object shall be the same as that of the old object prior to deallocation, up to the lesser of the new and old sizes. Any bytes in the new object beyond the size of the old object have indeterminate values.
If ptr is a null pointer, the realloc function behaves like the malloc function for the specified size. Otherwise, if ptr does not match a pointer earlier returned by a memory management function, or if the space has been deallocated by a call to the free or realloc function, the behavior is undefined. If size is nonzero and memory for the new object is not allocated, the old object is not deallocated. If size is zero and memory for the new object is not allocated, it is implementation-defined whether the old object is deallocated. If the old object is not deallocated, its value shall be unchanged.
What this means when you ask to reduce the size of a previously allocated object is:
The returned pointer might or might not be the same as the original pointer. (See discussion below.)
The C standard permits the portion of memory that is released to be reused for other allocations. Whether or not it is reused is up to the C implementation.
Whether the values in the released portion of memory are immediately overwritten or not is not specified by the C standard. Certainly the user of realloc may not rely on any behavior regarding that memory.
Discussion
When a memory allocation is reduced, it certainly seems “easy” for the memory allocation routines to simply return the same pointer while remembering that the released memory is free. However, memory allocation systems are fairly complex, so other factors may be involved. For example, hypothetically:
To support many small allocations without much overhead, a memory allocation system might create a pool of memory for one-to-four byte allocations, another pool for five-to-eight, another pool for eight-to-16, and a general pool for larger sizes. For the larger sizes, it might remember each allocation individually, customizing its size and managing them all with various data structures. For the smaller sizes, it might keep little more than a bitmap for each, with each bit indicating whether or not its corresponding four-byte (or eight or 16) region is allocated. In such a system, if you release eight bytes of a 16-byte allocation, the memory allocation software might move the data to something in the eight-byte pool.
In any memory allocation system, if you release just a few bytes at the end of an allocation, it might not be enough bytes to take advantage of—the data structures required to track the few bytes you released might be bigger than the few bytes. So it is not worthwhile to make them available for reuse. The memory allocation system just keeps them with the block, although it may remember the data in the block is actually a bit smaller than the space reserved for it.

Related

Can realloc fail when shrinking buffer? [duplicate]

If do the next:
int* array = malloc(10 * sizeof(int));
and them I use realloc:
array = realloc(array, 5 * sizeof(int));
On the second line (and only it), can it return NULL?
Yes, it can. There are no implementation guarantees on realloc(), and it can return a different pointer even when shrinking.
For example, if a particular implementation uses different pools for different object sizes, realloc() may actually allocate a new block in the pool for smaller objects and free the block in the pool for larger objects. Thus, if the pool for smaller objects is full, it will fail and return NULL.
Or it may simply decide it's better to move the block
I just used the following program to get size of actually allocated memory with glibc:
#include <stdlib.h>
#include <stdio.h>
int main()
{
int n;
for (n = 0; n <= 10; ++n)
{
void* array = malloc(n * sizeof(int));
size_t* a2 = (size_t*) array;
printf("%d -> %zu\n", n, a2[-1]);
}
}
and for n <= 6, it allocates 32 bytes, and for 7-10 it is 48.
So, if it shrank int[10] to int[5], the allocated size would shrink from 48 to 32, effectively giving 16 free bytes. Since (as it just has been noted) it won't allocate anything less than 32 bytes, those 16 bytes are lost.
If it moved the block elsewhere, the whole 48 bytes will be freed, and something could actually be put in there. Of course, that's just a science-fiction story and not a real implementation ;).
The most relevant quote from the C99 standard (7.20.3.4 The realloc function):
Returns
4 The realloc function returns a pointer to the new object (which may have the same value as a pointer to the old object), or a null pointer if the new object could not be allocated.
'May' is the key-word here. It doesn't mention any specific circumstances when that can happen, so you can't rely on any of them, even if they sound obvious at a first glance.
By the way, I think you could consider realloc() somewhat deprecated. If you'd take a look at C++, the newer memory allocation interfaces (new / delete and allocators) don't even support such a thing. They always expect you to allocate a new block. But that's just a loose comment.
The other answers have already nailed the question, but assuming you know the realloc call is a "trimming", you can wrap it with:
void *safe_trim(void *p, size_t n) {
void *p2 = realloc(p, n);
return p2 ? p2 : p;
}
and the return value will always point to an object of size n.
In any case, since the implementation of realloc knows the size of the object and can therefore determine that it's "trimming", it would be pathologically bad from a quality-of-implementation standpoint not to perform the above logic internally. But since realloc is not required to do this, you should do it yourself, either with the above wrapper or with analogous inline logic when you call realloc.
The language (and library) specification makes no such guarantee, just like it does not guarantee that a "trimming" realloc will preserve the pointer value.
An implementation might decide to implement realloc in the most "primitive" way: by doing an unconditional malloc for a new memory block, copying the data and free-ing the old block. Obviously, such implementation can fail in low-memory situations.
Don't count on it. The standard makes no such provision; it merely states "or a null pointer if the new object could not be allocated".
You'd be hard-pressed to find such an implementation, but according to the standard it would still be compliant.
I suspect there may be a theoretical possibility for failure in the scenario you describe.
Depending on the heap implementation, there may be no such a thing as trimming an existing allocation block. Instead a smaller block is allocated first, then the data is copied from the old one, and then it's freed.
For instance this may be the case with bucket-heap strategy (used by some popular heaps, such as tcmalloc).
A bit late, but there is at least one popular implementation which realloc() with a smaler size can fail: TCMalloc. (At least as far as i understand the code)
If you read the file tcmalloc.cc, in the function do_realloc_with_callback(), you will see that if you shrink enough (50% of alloced memory, otherwise it will be ignored), TCMalloc will alloc the new memory first (and possible fail) and then copy it and remove the old memory.
I do not copy the source code, because i am not sure if the copyrights (of TCMalloc and Stackoverflow) will allow that, but here is a link to the source (revision as at May 17, 2019).
realloc will not fails in shrinking the existing memory, so it will not return NULL. It can return NULL only if fails during expansion.
But shrinking can fail in some architecture, where realloc can be implemented in a different manner like allocating a smaller size memory separately and freeing the old memory to avoid fragmentation. In that case shrinking memory can return NULL. But its very rare implementation.
But its better to be in a safer side, to keep NULL checks after shrinking the memory also.

Why must malloc be used?

From what I understand, the malloc function takes a variable and allocates memory as asked. In this case, it will ask the compiler to prepare memory in order to fit the equivalence of twenty double variables. Is my way of understanding it correctly, and why must it be used?
double *q;
q=(double *)malloc(20*sizeof(double));
for (i=0;i<20; i++)
{
*(q+i)= (double) rand();
}
You don't have to use malloc() when:
The size is known at compile time, as in your example.
You are using C99 or C2011 with VLA (variable length array) support.
Note that malloc() allocates memory at runtime, not at compile time. The compiler is only involved to the extent that it ensures the correct function is called; it is malloc() that does the allocation.
Your example mentions 'equivalence of ten integers'. It is very seldom that 20 double occupy the same space as 10 int. Usually, 10 double will occupy the same space as 20 int (when sizeof(int) == 4 and sizeof(double) == 8, which is a very commonly found setting).
It's used to allocate memory at run-time rather than compile-time. So if your data arrays are based on some sort of input from the user, database, file, etc. then malloc must be used once the desired size is known.
The variable q is a pointer, meaning it stores an address in memory. malloc is asking the system to create a section of memory and return the address of that section of memory, which is stored in q. So q points to the starting location of the memory you requested.
Care must be taken not to alter q unintentionally. For instance, if you did:
q = (double *)malloc(20*sizeof(double));
q = (double *)malloc(10*sizeof(double));
you will lose access to the first section of 20 double's and introduce a memory leak.
When you use malloc you are asking the system "Hey, I want this many bytes of memory" and then he will either say "Sorry, I'm all out" or "Ok! Here is an address to the memory you wanted. Don't lose it".
It's generally a good idea to put big datasets in the heap (where malloc gets your memory from) and a pointer to that memory on the stack (where code execution takes place). This becomes more important on embedded platforms where you have limited memory. You have to decide how you want to divvy up the physical memory between the stack and heap. Too much stack and you can't dynamically allocate much memory. Too little stack and you can function call your way right out of it (also known as a stack overflow :P)
As the others said, malloc is used to allocate memory. It is important to note that malloc will allocate memory from the heap, and thus the memory is persistent until it is free'd. Otherwise, without malloc, declaring something like double vals[20] will allocate memory on the stack. When you exit the function, that memory is popped off of the stack.
So for example, say you are in a function and you don't care about the persistence of values. Then the following would be suitable:
void some_function() {
double vals[20];
for(int i = 0; i < 20; i++) {
vals[i] = (double)rand();
}
}
Now if you have some global structure or something that stores data, that has a lifetime longer than that of just the function, then using malloc to allocate that memory from the heap is required (alternatively, you can declare it as a global variable, and the memory will be preallocated for you).
In you example, you could have declared double q[20]; without the malloc and it would work.
malloc is a standard way to get dynamically allocated memory (malloc is often built above low-level memory acquisition primitives like mmap on Linux).
You want to get dynamically allocated memory resources, notably when the size of the allocated thing (here, your q pointer) depends upon runtime parameters (e.g. depends upon input). The bad alternative would be to allocate all statically, but then the static size of your data is a strong built-in limitation, and you don't like that.
Dynamic resource allocation enables you to run the same program on a cheap tablet (with half a gigabyte of RAM) and an expensive super-computer (with terabytes of RAM). You can allocate different size of data.
Don't forget to test the result of malloc; it can fail by returning NULL. At the very least, code:
int* q = malloc (10*sizeof(int));
if (!q) {
perror("q allocation failed");
exit(EXIT_FAILURE);
};
and always initialize malloc-ed memory (you could prefer using calloc which zeroes the allocated memory).
Don't forget to later free the malloc-ed memory. On Linux, learn about using valgrind. Be scared of memory leaks and dangling pointers. Recognize that the liveness of some data is a non-modular property of the entire program. Read about garbage collection!, and consider perhaps using Boehm's conservative garbage collector (by calling GC_malloc instead of malloc).
You use malloc() to allocate memory dynamically in C. (Allocate the memory at the run time)
You use it because sometimes you don't know how much memory you'll use when you write your program.
You don't have to use it when you know thow many elements the array will hold at compile time.
Another important thing to notice that if you want to return an array from a function, you will want to return an array which was not defined inside the function on the stack. Instead, you'll want to dynamically allocate an array (on the heap) and return a pointer to this block:
int *returnArray(int n)
{
int i;
int *arr = (int *)malloc(sizeof(int) * n);
if (arr == NULL)
{
return NULL;
}
//...
//fill the array or manipulate it
//...
return arr; //return the pointer
}

Can realloc fail (return NULL) when trimming?

If do the next:
int* array = malloc(10 * sizeof(int));
and them I use realloc:
array = realloc(array, 5 * sizeof(int));
On the second line (and only it), can it return NULL?
Yes, it can. There are no implementation guarantees on realloc(), and it can return a different pointer even when shrinking.
For example, if a particular implementation uses different pools for different object sizes, realloc() may actually allocate a new block in the pool for smaller objects and free the block in the pool for larger objects. Thus, if the pool for smaller objects is full, it will fail and return NULL.
Or it may simply decide it's better to move the block
I just used the following program to get size of actually allocated memory with glibc:
#include <stdlib.h>
#include <stdio.h>
int main()
{
int n;
for (n = 0; n <= 10; ++n)
{
void* array = malloc(n * sizeof(int));
size_t* a2 = (size_t*) array;
printf("%d -> %zu\n", n, a2[-1]);
}
}
and for n <= 6, it allocates 32 bytes, and for 7-10 it is 48.
So, if it shrank int[10] to int[5], the allocated size would shrink from 48 to 32, effectively giving 16 free bytes. Since (as it just has been noted) it won't allocate anything less than 32 bytes, those 16 bytes are lost.
If it moved the block elsewhere, the whole 48 bytes will be freed, and something could actually be put in there. Of course, that's just a science-fiction story and not a real implementation ;).
The most relevant quote from the C99 standard (7.20.3.4 The realloc function):
Returns
4 The realloc function returns a pointer to the new object (which may have the same value as a pointer to the old object), or a null pointer if the new object could not be allocated.
'May' is the key-word here. It doesn't mention any specific circumstances when that can happen, so you can't rely on any of them, even if they sound obvious at a first glance.
By the way, I think you could consider realloc() somewhat deprecated. If you'd take a look at C++, the newer memory allocation interfaces (new / delete and allocators) don't even support such a thing. They always expect you to allocate a new block. But that's just a loose comment.
The other answers have already nailed the question, but assuming you know the realloc call is a "trimming", you can wrap it with:
void *safe_trim(void *p, size_t n) {
void *p2 = realloc(p, n);
return p2 ? p2 : p;
}
and the return value will always point to an object of size n.
In any case, since the implementation of realloc knows the size of the object and can therefore determine that it's "trimming", it would be pathologically bad from a quality-of-implementation standpoint not to perform the above logic internally. But since realloc is not required to do this, you should do it yourself, either with the above wrapper or with analogous inline logic when you call realloc.
The language (and library) specification makes no such guarantee, just like it does not guarantee that a "trimming" realloc will preserve the pointer value.
An implementation might decide to implement realloc in the most "primitive" way: by doing an unconditional malloc for a new memory block, copying the data and free-ing the old block. Obviously, such implementation can fail in low-memory situations.
Don't count on it. The standard makes no such provision; it merely states "or a null pointer if the new object could not be allocated".
You'd be hard-pressed to find such an implementation, but according to the standard it would still be compliant.
I suspect there may be a theoretical possibility for failure in the scenario you describe.
Depending on the heap implementation, there may be no such a thing as trimming an existing allocation block. Instead a smaller block is allocated first, then the data is copied from the old one, and then it's freed.
For instance this may be the case with bucket-heap strategy (used by some popular heaps, such as tcmalloc).
A bit late, but there is at least one popular implementation which realloc() with a smaler size can fail: TCMalloc. (At least as far as i understand the code)
If you read the file tcmalloc.cc, in the function do_realloc_with_callback(), you will see that if you shrink enough (50% of alloced memory, otherwise it will be ignored), TCMalloc will alloc the new memory first (and possible fail) and then copy it and remove the old memory.
I do not copy the source code, because i am not sure if the copyrights (of TCMalloc and Stackoverflow) will allow that, but here is a link to the source (revision as at May 17, 2019).
realloc will not fails in shrinking the existing memory, so it will not return NULL. It can return NULL only if fails during expansion.
But shrinking can fail in some architecture, where realloc can be implemented in a different manner like allocating a smaller size memory separately and freeing the old memory to avoid fragmentation. In that case shrinking memory can return NULL. But its very rare implementation.
But its better to be in a safer side, to keep NULL checks after shrinking the memory also.

Freeing portions of dynamically allocated blocks?

I was curious whether there exists a dynamic memory allocation system that allows the programmer to free part of an allocated block.
For example:
char* a = malloc (40);
//b points to the split second half of the block, or to NULL if it's beyond the end
//a points to a area of 10 bytes
b = partial_free (a+10, /*size*/ 10)
Thoughts on why this is wise/unwise/difficult? Ways to do this?
Seems to me like it could be useful.
Thanks!
=====edit=====
after some research, it seems that the bootmem allocator for the linux kernel allows something similar to this operation with the bootmem_free call. So, I'm curious -- why is it that the bootmem allocator allows this, but ANSI C does not?
No there is no such function which allows parital freeing of memory.
You could however use realloc() to resize memory.
From the c standard:
7.22.3.5 The realloc function
#include <stdlib.h>
void *realloc(void *ptr, size_t size);
The realloc function deallocates the old object pointed to by ptr and returns a pointer to a new object that has the size specified by size. The contents of the new object shall be the same as that of the old object prior to deallocation, up to the lesser of the new and old sizes. Any bytes in the new object beyond the size of the old object have indeterminate values.
There is no ready-made function for this, but doing this isn't impossible. Firstly, there is realloc() . realloc takes a pointer to a block of memory and resizes the allocation to the size specified.
Now, if you have allocated some memory:
char * tmp = malloc(2048);
and you intend to deallocate the first, 1 K of memory, you may do:
tmp = realloc(foo, 2048-1024);
However, the problem in this case is that you cannot be certain that tmp will remain unchanged. Since, the function might just deallocate the entire 2K memory and move it elsewhere.
Now I'm not sure about the exact implementation of realloc, but from what I understand, the code:
myptr = malloc( x - y );
actually mallocs a new memory buffer of size x-y, then it copies the bytes that fit using memcpy and finally frees the original allocated memory.
This may create some potential problems. For example, the new reallocated memory may be located at a different address, so any past pointers you may have may become invalidated. Resulting in undefined runtime errors, segmentation faults and general debugging hell. So I would try to avoid resorting to this.
Firstly, I cannot think of any situation where you would be likely to need such a thing (when there exists realloc to increase/decrease the memory as mentioned in the answers).
I would like to add another thing. In whatever implementations I have seen of the malloc subsystem (which I admit is not a lot), malloc and free are implemented to be dependent on something called as the prefix byte(s). So whatever address is returned to you by malloc, internally the malloc subsystem will allocate some additional byte(s) of memory prior to the address returned to you, to store sanity check information which includes number of allocated bytes and possible what allocation policy you use (if your OS supports multiple mem allocation policies) etc. When you say something like free (x bytes), the malloc subsystem goes back to peek back into the prefix byte to sanity check and only if it finds the prefix in place does the free successfully happen. Therefore, it will not allow you to free some number of blocks starting in between.

Instead of just using free() and having the pointer pointing some new block, how to really empty the previously-pointed-at memory block?

I am trying to free dynamically allocated memory using free(), but I found that what it does is to have the argument pointer point to some new location, and leaving the previously-pointed-at location as it was, the memory is not cleared. And if I use malloc again, the pointer may point to this messy block, and it's already filled with garbage, which is really annoying..
I'm kinda new to C and I think delete[] in c++ doesn't have this problem. Any advise?
Thanks
By free the memory is just released from use. It is released from being allocated to you. it is not explicitly cleared. Some old contents might be present at those memory locations.
To avoid this, there are two solutions.
Solution 1:
You will need to do a memset after allocating memory using malloc.
Code Example:
unsigned int len = 20; // len is the length of boo
char* bar = 0;
bar= (char *)malloc(len);
memset(bar, 0, len);
Solution 2:
Or use, calloc() which initiliazes memory to 0 by default.
Code Example:
int *pData = 0;
int i = 10;
pData = (int*) calloc (i,sizeof(int));
I think delete[] in c++ doesn't have this problem.
No
It behaves exactly this same way. Unless you explicitly set the pointer to 0 the delete'd pointer will not be pointing to 0. So do always set the pointer to 0 after you delete it.
When should you use malloc over calloc or vice versa?
Since calloc sets the allocated memory to 0 this may take a little time, so you may probably want to use malloc() if that performance is an issue.(Ofcourse One most profile their usage to see if this really is a problem)
If initializing the memory is more important, use calloc() as it does that explicitly for you.
Also, some OS like Linux have an Lazy Allocation memory model wherein the returned memory address is a virtual address and the actual allocation only happens at run-time. The OS assumes that it will be able to provide this allocation at Run-Time.
The memory allocated by malloc is not backed by real memory until the program actually touches it.
While, since calloc initializes the memory to 0 you can be assured that the OS has already backed the allocation with actual RAM (or swap).
How about realloc?
Yes, similar behavior to malloc.
Excerpt From the documentation:
void * realloc ( void * ptr, size_t size );
Reallocate memory block
The size of the memory block pointed to by the ptr parameter is changed to the size bytes, expanding or reducing the amount of memory available in the block.
The function may move the memory block to a new location, in which case the new location is returned. The content of the memory block is preserved up to the lesser of the new and old sizes, even if the block is moved.If the new size is larger, the value of the newly allocated portion is indeterminate.
In case that ptr is NULL, the function behaves exactly as malloc, assigning a new block of size bytes and returning a pointer to the beginning of it.
In case that the size is 0, the memory previously allocated in ptr is deallocated as if a call to free was made, and a NULL pointer is returned.
You can use calloc( ) instead of malloc( ) to clear the allocated memory to zero.
Why is having newly allocated memory filled with garbage "really annoying"? If you allocate memory, presumably it's because you're going to use it for something -- which means you have to store some meaningful value into it before attempting to read it. In most cases, in well-written code, there's no reason to care what's in newly allocated memory.
If you happen to have a requirement for a newly allocated block of memory you can call memset after calling malloc, or you can use calloc instead of malloc. But consider carefully whether there's any real advantage in doing so. If you're actually going to use those all-bits-zero values (i.e., if all-bits-zero happens to be the "meaningful value" I mentioned above), go ahead and clear the block. (But keep in mind that the language doesn't guarantee that either a null pointer or a floating-point 0.0 is represented as all-bits-zero, though it is in most implementations they are.)
And free() doesn't "have the argument pointer point to some new location". free(ptr) causes the memory pointed to by ptr to be made available for future allocation. It doesn't change the contents of the pointer object ptr itself (though the address stored in ptr does become invalid).

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