This question already has answers here:
Getting a stack overflow exception when declaring a large array
(8 answers)
Closed 5 years ago.
Hi all I am really new to C (just started this week), and want to make sure that I am not looking down the wrong rabbit home and hoping to perhaps get pointed to the right rabbit hole.
I create a struct:
#define MAX 64
#define ARRAY_SIZE 2048
struct object {
int q, c, p;
char name[MAX]; //Stores string up to 63 characters
char arr[ARRAY_SIZE][MAX]; // Creates an array of 2048 cells with string of length 63 max
};
int main(){
...
...
int variable = 30;
struct object l[variable]; //This is where the crash happens. But only when either variable is too large (for instance works on 15 just fine, but anything over 20 it crashes), or when Array_SIZE is too larger, for instance works fine with 1024 but 2048 crashes.
...
...
}
The error I get on crash is the following: Process returned -1073741571 (0xC00000FD) in the cmd window. And the following in the debugger in the IDE:
Program received signal SIGSEGV, Segmentation fault.
[Inferior 1 (process 12120) exited with code 030000000375]
Am I doing something obviously wrong with how I declare an array of structs? Why would large numbers not work but lower numebrs work?
Does the above error indicate I am accessing something out of bounds somewhere? Ive been up and down the code and cant seem to find any reason why larger numbers dont work and lower ones do. My memory footprint doesnt seem to be the issue, just a few megs of memory.
I need help with what to look for (I cant find any instances of accessing anything out of bounds, so I get the feeling Im not chasing the right rabbit and need to look for something else)? Or maybe Im doing something illegal for C without knowing it?
I think your program crashes because you statically allocate too much memory on the stack.
Try using the malloc or calloc function. It dynamically allocates memory on the heap instead e.g. :
struct object *l = malloc(variable*sizeof(struct object));
Don't forget to free it afterwards using the free function.
free(l);
You have a memory size problem, try to increase the memory size for your program.
And if you want to use big array size and so allocate a lot of memory, you shouldn't allocate statically but dynamically.
So you should use malloc
typedef struct object {
int q, c, p;
char name[MAX]; //Stores string up to 63 characters
char arr[ARRAY_SIZE][MAX];
} myObject ;
int variable = 30;
myObject *l = malloc(sizeof(myObject) * variable);
I do not advice you to declare an array of 2048 statically, so you should initiate your struct with a function.
typedef struct object {
int q, c, p;
char name[MAX]; //Stores string up to 63 characters
char *arr[MAX];
} myObject ;
myObject *createNewObject() {
myObject *toReturn = malloc(sizeof(myObject) * variable);
if (toReturn == NULL)
return NULL;
toReturn->arr = malloc(sizeof(char) * ARRAY_SIZE);
return (toReturn);
}
void freeMyObject(myObject *objectToFree)
{
if (objectToFree && objectToFree->arr =! NULL)
free(objectToFree->arr)
if (objectToFree)
free(objectToFree)
}
void main()
{
myObject *myNewObj = createNewObject()
// You can do some stuff with myNewObj but always verify the pointers
freeMyObject(myNewObj);
}
You should also debug with valgrind when you works with malloc, so you don't have memory loss or problems.
Hope I helped
Well the problem you had is - you have used automatic memory allocation. Due to constraint of size of automatic storage your program crashed - you asked for more than you should.
So what is the solution?
Static memory allocation:
Solution being
static struct object l[variable];
Dynamic memory allocation
struct object *ptr = malloc( sizeof *ptr * variable);
The storage of these allocation is different from automatic variables - so free from the size constraint . In this case you have to free the dynamically allocated memory. That's why you will get around the problem you have.
Statically allocate is not a confusion free term. All these types of variable will have different scope - lifetime. Standard never mentions about stack or heap . It is the implementation that follow these to store automatically allocated memory and dynamically allocated memory.
This question already has answers here:
How is the array stored in memory?
(4 answers)
How to access a local variable from a different function using pointers?
(10 answers)
Closed 6 years ago.
I am trying to implement a simple program using a header file where a function in the header file accepts an int array and returns an int array too.
In header.h:
int* point(int a[]);
In header.c:
#include<stdio.h>
#include "header.h"
int* point(int a[])
{
printf("In the point function\n");
int array[4],i;
for(int i=0;i<4;i++)
{
printf("%dth Iteration\n",i);
array[i]=a[i];
}
return array;
}
In test.c:
#include<stdio.h>
#include "header.h"
void main()
{
int *array,i;
int a[]={1,2,3,4};
printf("calling point function\n");
array=point(a);
printf("Back in the main function\n");
for(i=0;i<4;i++)
{
//SEGMENTATION FAULT HERE
printf("%d\n",array[i]);
}
}
I am getting a segmentation fault at the print loop in test.c.
You cannot return arrays from functions. When point() returns, the local array within this function goes out of scope. This array is created on the stack, and will get destroyed once the function finishes returning. All memory associated with it is discarded, and the returned pointer points to a position on the stack that doesn't exist anymore. You need to instead allocate a pointer on the heap, and return that instead. This allows array to be shared across your program.
Instead of:
int array[4];
you need to dynamically allocate a pointer using malloc():
int *array = malloc(4 * sizeof(*array)); /* or sizeof(int) */
if (array == NULL) {
/* handle exit */
}
malloc() allocates requested memory on the heap, and returns a void* pointer to it.
Note: malloc() can return NULL when unsuccessful, so it needs to be checked always. You also need to free() any memory previously allocated by malloc(). You also don't need to cast return of malloc().
Another thing to point out is using the magic number 4 all over your program. This should really be calculated using sizeof(a)/sizeof(a[0]).
You can declare this as a size_t variable in your main():
size_t n = sizeof(a)/sizeof(a[0]);
Or you can use a macro:
#define ARRAYSIZE(arr) (sizeof(arr) / sizeof(arr[0]))
And simply call ARRAYSIZE(a) everytime you want the size of the array.
The issue has to do with the scope of the array variable that you're returning in your method. Right now you're returning array, a local variable defined in the method, point. However, once point is finished executing, all local variables within the function frame, including array will be discarded from main memory. So even though you still get a memory address from point, there's no telling what could be at that memory address. Therefore, treating array as an int array when printing out its elements will lead to a segmentation fault.
My suggestion to fix this is to allocate memory from the heap using malloc so that array lasts outside the frame of point. The solution should look like this,
int* point(int a[])
{
printf("In the point function\n");
int *array = (int *) malloc(4 * sizeof(int)); //dynamically allocate memory for 4 integers
int i;
for(i=0;i<4;i++)
{
printf("%dth Iteration\n",i);
array[i]=a[i];
}
return array;
}
You could either define array[] as a global variable, or dynamically allocate memory for it as mentioned in the above comments using malloc().
Since array[] is allocated in the function point(), it gets deleted once the function exits. Hence, a reference to the returned pointer causes a segmentation fault.
This question already has answers here:
Initially mallocate 0 elements to later reallocate and measure size
(2 answers)
Closed 6 years ago.
Below code is just an example which I use to come to the point later:
/* Extract digits from an integer and store in an array in reverse order*/
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int oct=316;
int* array=malloc(0*sizeof(int*)); //Step 1
int* temp;
size_t size=0;
while(oct/10>0 || oct%10>0){
temp=realloc(array,++size*sizeof(int)); // Step2 requires Step1
if(temp==NULL){
printf("Can't reallocate memory");
exit(-1);
}
else{
array=temp;
}
array[size-1]=oct%10;
oct/=10;
}
for(int i=0;i<size;i++)
printf("%d\n",array[i]);
return 0;
}
The realloc reference [1] states :
Reallocates the given area of memory. It must be previously allocated
by malloc(), calloc(), realloc()...
Initially I compiled the code without step1 and on running it I got a segmentation fault. Then I included step1, now the code compiles fine. I don't want to allocate some space without finding an integer to store so I have used a size of zero with malloc. However malloc reference [2] states :
If size is zero, the return value depends on the particular library
implementation (it may or may not be a null pointer), but the returned
pointer shall not be dereferenced.
Now, I doubt my implementation is portable. How can I work around this issue?
I tried doing int* array=NULL; but I got a segmentation fault.
You can initiate array to NULL. In the man page for realloc(void *ptr, size_t size) you can read:
If ptr is NULL, then the call is equivalent to malloc(size)
Moreover realloc does not change its parameter. It returns pointer to newly allocated/reallocated space. So, for reallocating you shall use a code like:
array = realloc(array,(++size)*sizeof(int));
if (array == NULL) { some error; }
I've allocated an "array" of mystruct of size n like this:
if (NULL == (p = calloc(sizeof(struct mystruct) * n,1))) {
/* handle error */
}
Later on, I only have access to p, and no longer have n. Is there a way to determine the length of the array given just the pointer p?
I figure it must be possible, since free(p) does just that. I know malloc() keeps track of how much memory it has allocated, and that's why it knows the length; perhaps there is a way to query for this information? Something like...
int length = askMallocLibraryHowMuchMemoryWasAlloced(p) / sizeof(mystruct)
I know I should just rework the code so that I know n, but I'd rather not if possible. Any ideas?
No, there is no way to get this information without depending strongly on the implementation details of malloc. In particular, malloc may allocate more bytes than you request (e.g. for efficiency in a particular memory architecture). It would be much better to redesign your code so that you keep track of n explicitly. The alternative is at least as much redesign and a much more dangerous approach (given that it's non-standard, abuses the semantics of pointers, and will be a maintenance nightmare for those that come after you): store the lengthn at the malloc'd address, followed by the array. Allocation would then be:
void *p = calloc(sizeof(struct mystruct) * n + sizeof(unsigned long int),1));
*((unsigned long int*)p) = n;
n is now stored at *((unsigned long int*)p) and the start of your array is now
void *arr = p+sizeof(unsigned long int);
Edit: Just to play devil's advocate... I know that these "solutions" all require redesigns, but let's play it out.
Of course, the solution presented above is just a hacky implementation of a (well-packed) struct. You might as well define:
typedef struct {
unsigned int n;
void *arr;
} arrInfo;
and pass around arrInfos rather than raw pointers.
Now we're cooking. But as long as you're redesigning, why stop here? What you really want is an abstract data type (ADT). Any introductory text for an algorithms and data structures class would do it. An ADT defines the public interface of a data type but hides the implementation of that data type. Thus, publicly an ADT for an array might look like
typedef void* arrayInfo;
(arrayInfo)newArrayInfo(unsignd int n, unsigned int itemSize);
(void)deleteArrayInfo(arrayInfo);
(unsigned int)arrayLength(arrayInfo);
(void*)arrayPtr(arrayInfo);
...
In other words, an ADT is a form of data and behavior encapsulation... in other words, it's about as close as you can get to Object-Oriented Programming using straight C. Unless you're stuck on a platform that doesn't have a C++ compiler, you might as well go whole hog and just use an STL std::vector.
There, we've taken a simple question about C and ended up at C++. God help us all.
keep track of the array size yourself; free uses the malloc chain to free the block that was allocated, which does not necessarily have the same size as the array you requested
Just to confirm the previous answers: There is no way to know, just by studying a pointer, how much memory was allocated by a malloc which returned this pointer.
What if it worked?
One example of why this is not possible. Let's imagine the code with an hypothetic function called get_size(void *) which returns the memory allocated for a pointer:
typedef struct MyStructTag
{ /* etc. */ } MyStruct ;
void doSomething(MyStruct * p)
{
/* well... extract the memory allocated? */
size_t i = get_size(p) ;
initializeMyStructArray(p, i) ;
}
void doSomethingElse()
{
MyStruct * s = malloc(sizeof(MyStruct) * 10) ; /* Allocate 10 items */
doSomething(s) ;
}
Why even if it worked, it would not work anyway?
But the problem of this approach is that, in C, you can play with pointer arithmetics. Let's rewrite doSomethingElse():
void doSomethingElse()
{
MyStruct * s = malloc(sizeof(MyStruct) * 10) ; /* Allocate 10 items */
MyStruct * s2 = s + 5 ; /* s2 points to the 5th item */
doSomething(s2) ; /* Oops */
}
How get_size is supposed to work, as you sent the function a valid pointer, but not the one returned by malloc. And even if get_size went through all the trouble to find the size (i.e. in an inefficient way), it would return, in this case, a value that would be wrong in your context.
Conclusion
There are always ways to avoid this problem, and in C, you can always write your own allocator, but again, it is perhaps too much trouble when all you need is to remember how much memory was allocated.
Some compilers provide msize() or similar functions (_msize() etc), that let you do exactly that
May I recommend a terrible way to do it?
Allocate all your arrays as follows:
void *blockOfMem = malloc(sizeof(mystruct)*n + sizeof(int));
((int *)blockofMem)[0] = n;
mystruct *structs = (mystruct *)(((int *)blockOfMem) + 1);
Then you can always cast your arrays to int * and access the -1st element.
Be sure to free that pointer, and not the array pointer itself!
Also, this will likely cause terrible bugs that will leave you tearing your hair out. Maybe you can wrap the alloc funcs in API calls or something.
malloc will return a block of memory at least as big as you requested, but possibly bigger. So even if you could query the block size, this would not reliably give you your array size. So you'll just have to modify your code to keep track of it yourself.
For an array of pointers you can use a NULL-terminated array. The length can then determinate like it is done with strings. In your example you can maybe use an structure attribute to mark then end. Of course that depends if there is a member that cannot be NULL. So lets say you have an attribute name, that needs to be set for every struct in your array you can then query the size by:
int size;
struct mystruct *cur;
for (cur = myarray; cur->name != NULL; cur++)
;
size = cur - myarray;
Btw it should be calloc(n, sizeof(struct mystruct)) in your example.
Other have discussed the limits of plain c pointers and the stdlib.h implementations of malloc(). Some implementations provide extensions which return the allocated block size which may be larger than the requested size.
If you must have this behavior you can use or write a specialized memory allocator. This simplest thing to do would be implementing a wrapper around the stdlib.h functions. Some thing like:
void* my_malloc(size_t s); /* Calls malloc(s), and if successful stores
(p,s) in a list of handled blocks */
void my_free(void* p); /* Removes list entry and calls free(p) */
size_t my_block_size(void* p); /* Looks up p, and returns the stored size */
...
really your question is - "can I find out the size of a malloc'd (or calloc'd) data block". And as others have said: no, not in a standard way.
However there are custom malloc implementations that do it - for example http://dmalloc.com/
I'm not aware of a way, but I would imagine it would deal with mucking around in malloc's internals which is generally a very, very bad idea.
Why is it that you can't store the size of memory you allocated?
EDIT: If you know that you should rework the code so you know n, well, do it. Yes it might be quick and easy to try to poll malloc but knowing n for sure would minimize confusion and strengthen the design.
One of the reasons that you can't ask the malloc library how big a block is, is that the allocator will usually round up the size of your request to meet some minimum granularity requirement (for example, 16 bytes). So if you ask for 5 bytes, you'll get a block of size 16 back. If you were to take 16 and divide by 5, you would get three elements when you really only allocated one. It would take extra space for the malloc library to keep track of how many bytes you asked for in the first place, so it's best for you to keep track of that yourself.
This is a test of my sort routine. It sets up 7 variables to hold float values, then assigns them to an array, which is used to find the max value.
The magic is in the call to myMax:
float mmax = myMax((float *)&arr,(int) sizeof(arr)/sizeof(arr[0]));
And that was magical, wasn't it?
myMax expects a float array pointer (float *) so I use &arr to get the address of the array, and cast it as a float pointer.
myMax also expects the number of elements in the array as an int. I get that value by using sizeof() to give me byte sizes of the array and the first element of the array, then divide the total bytes by the number of bytes in each element. (we should not guess or hard code the size of an int because it's 2 bytes on some system and 4 on some like my OS X Mac, and could be something else on others).
NOTE:All this is important when your data may have a varying number of samples.
Here's the test code:
#include <stdio.h>
float a, b, c, d, e, f, g;
float myMax(float *apa,int soa){
int i;
float max = apa[0];
for(i=0; i< soa; i++){
if (apa[i]>max){max=apa[i];}
printf("on i=%d val is %0.2f max is %0.2f, soa=%d\n",i,apa[i],max,soa);
}
return max;
}
int main(void)
{
a = 2.0;
b = 1.0;
c = 4.0;
d = 3.0;
e = 7.0;
f = 9.0;
g = 5.0;
float arr[] = {a,b,c,d,e,f,g};
float mmax = myMax((float *)&arr,(int) sizeof(arr)/sizeof(arr[0]));
printf("mmax = %0.2f\n",mmax);
return 0;
}
In uClibc, there is a MALLOC_SIZE macro in malloc.h:
/* The size of a malloc allocation is stored in a size_t word
MALLOC_HEADER_SIZE bytes prior to the start address of the allocation:
+--------+---------+-------------------+
| SIZE |(unused) | allocation ... |
+--------+---------+-------------------+
^ BASE ^ ADDR
^ ADDR - MALLOC_HEADER_SIZE
*/
/* The amount of extra space used by the malloc header. */
#define MALLOC_HEADER_SIZE \
(MALLOC_ALIGNMENT < sizeof (size_t) \
? sizeof (size_t) \
: MALLOC_ALIGNMENT)
/* Set up the malloc header, and return the user address of a malloc block. */
#define MALLOC_SETUP(base, size) \
(MALLOC_SET_SIZE (base, size), (void *)((char *)base + MALLOC_HEADER_SIZE))
/* Set the size of a malloc allocation, given the base address. */
#define MALLOC_SET_SIZE(base, size) (*(size_t *)(base) = (size))
/* Return base-address of a malloc allocation, given the user address. */
#define MALLOC_BASE(addr) ((void *)((char *)addr - MALLOC_HEADER_SIZE))
/* Return the size of a malloc allocation, given the user address. */
#define MALLOC_SIZE(addr) (*(size_t *)MALLOC_BASE(addr))
malloc() stores metadata regarding space allocation before 8 bytes from space actually allocated. This could be used to determine space of buffer. And on my x86-64 this always return multiple of 16. So if allocated space is multiple of 16 (which is in most cases) then this could be used:
Code
#include <stdio.h>
#include <malloc.h>
int size_of_buff(void *buff) {
return ( *( ( int * ) buff - 2 ) - 17 ); // 32 bit system: ( *( ( int * ) buff - 1 ) - 17 )
}
void main() {
char *buff = malloc(1024);
printf("Size of Buffer: %d\n", size_of_buff(buff));
}
Output
Size of Buffer: 1024
This is my approach:
#include <stdio.h>
#include <stdlib.h>
typedef struct _int_array
{
int *number;
int size;
} int_array;
int int_array_append(int_array *a, int n)
{
static char c = 0;
if(!c)
{
a->number = NULL;
a->size = 0;
c++;
}
int *more_numbers = NULL;
a->size++;
more_numbers = (int *)realloc(a->number, a->size * sizeof(int));
if(more_numbers != NULL)
{
a->number = more_numbers;
a->number[a->size - 1] = n;
}
else
{
free(a->number);
printf("Error (re)allocating memory.\n");
return 1;
}
return 0;
}
int main()
{
int_array a;
int_array_append(&a, 10);
int_array_append(&a, 20);
int_array_append(&a, 30);
int_array_append(&a, 40);
int i;
for(i = 0; i < a.size; i++)
printf("%d\n", a.number[i]);
printf("\nLen: %d\nSize: %d\n", a.size, a.size * sizeof(int));
free(a.number);
return 0;
}
Output:
10
20
30
40
Len: 4
Size: 16
If your compiler supports VLA (variable length array), you can embed the array length into the pointer type.
int n = 10;
int (*p)[n] = malloc(n * sizeof(int));
n = 3;
printf("%d\n", sizeof(*p)/sizeof(**p));
The output is 10.
You could also choose to embed the information into the allocated memory yourself with a structure including a flexible array member.
struct myarray {
int n;
struct mystruct a[];
};
struct myarray *ma =
malloc(sizeof(*ma) + n * sizeof(struct mystruct));
ma->n = n;
struct mystruct *p = ma->a;
Then to recover the size, you would subtract the offset of the flexible member.
int get_size (struct mystruct *p) {
struct myarray *ma;
char *x = (char *)p;
ma = (void *)(x - offsetof(struct myarray, a));
return ma->n;
}
The problem with trying to peek into heap structures is that the layout might change from platform to platform or from release to release, and so the information may not be reliably obtainable.
Even if you knew exactly how to peek into the meta information maintained by your allocator, the information stored there may have nothing to do with the size of the array. The allocator simply returned memory that could be used to fit the requested size, but the actual size of the memory may be larger (perhaps even much larger) than the requested amount.
The only reliable way to know the information is to find a way to track it yourself.
This question already has answers here:
Determine size of dynamically allocated memory in C
(15 answers)
Closed 3 years ago.
I have faced some problem in this case can you please your ideas.
main()
{
char *p=NULL;
p=(char *)malloc(2000 * sizeof(char));
printf("size of p = %d\n",sizeof (p));
}
In this program Its print the 4 that (char *) value,but i need how many bytes allocated for
that.
You could also implement a wrapper for malloc and free to add tags (like allocated size and other meta information) before the pointer returned by malloc. This is in fact the method that a c++ compiler tags objects with references to virtual classes.
Here is one working example:
#include <stdlib.h>
#include <stdio.h>
void * my_malloc(size_t s)
{
size_t * ret = malloc(sizeof(size_t) + s);
*ret = s;
return &ret[1];
}
void my_free(void * ptr)
{
free( (size_t*)ptr - 1);
}
size_t allocated_size(void * ptr)
{
return ((size_t*)ptr)[-1];
}
int main(int argc, const char ** argv) {
int * array = my_malloc(sizeof(int) * 3);
printf("%u\n", allocated_size(array));
my_free(array);
return 0;
}
The advantage of this method over a structure with size and pointer
struct pointer
{
size_t size;
void *p;
};
is that you only need to replace the malloc and free calls. All other pointer operations require no refactoring.
There is no portable way but for windows:
#include <stdio.h>
#include <malloc.h>
#if defined( _MSC_VER ) || defined( __int64 ) /* for VisualC++ or MinGW/gcc */
#define howmanybytes(ptr) ((unsigned long)_msize(ptr))
#else
#error no known way
#endif
int main()
{
char *x=malloc(1234);
printf( "%lu", howmanybytes(x) );
return 0;
}
Although it may be possible that some libraries allows you to determine the size of an allocated buffer, it wouldn't be a standard C function and you should be looking at your library's own documentations for this.
However, if there are many places that you need to know the size of your allocated memory, the cleanest way you could do it is to keep the size next to the pointer. That is:
struct pointer
{
size_t size;
void *p;
};
Then every time you malloc the pointer, you write down the size in the size field also. The problem with this method however is that you have to cast the pointer every time you use it. If you were in C++, I would have suggested using template classes. However, in this case also it's not hard, just create as many structs as the types you have. So for example
struct charPtr
{
size_t size;
char *p;
};
struct intPtr
{
size_t size;
int *p;
};
struct objectPtr
{
size_t size;
struct object *p;
};
Given similar names, once you define the pointer, you don't need extra effort (such as casting) to access the array. An example of usage is:
struct intPtr array;
array.p = malloc(1000 * sizeof *array.p);
array.size = array.p?1000:0;
...
for (i = 0; i < array.size; ++i)
printf("%s%d", i?" ":"", array.p[i]);
printf("\n");
It is impossible to know how much memory was allocated by just the pointer. doing sizeof (p) will get the size of the pointer variable p which it takes at compile time, and which is the size of the pointer. That is, the memory the pointer variable takes to store the pointer variable p. Inside p the starting address of the memory block is stored.
Once you allocate some memory with malloc it will return the starting address of the memory block, but the end of the block cannot be found from it, as there is no terminator for a block. You define the end of the block therefore you need to identify it by any means, so store it somewhere. Therefore you need to preserve the block length somewhere to know where the block which is pointed to by p ends.
Note: Although the memory allocation structure keeps track of allocated and unallocated blocks, therefore we can know the allocated memory block length from these structures, but these structures are not available to be used by the users, unless any library function provides them. Therefore a code using such feature is not portable (pointed by #Rudy Velthuis) . Therefore it is the best to keep track of the structure yourself.
You need to keep track of it in a variable if you want to know it for later:
char *p = NULL;
int sizeofp = 2000*sizeof(char);
p = (char *)malloc(sizeofp);
printf("size of p = %d\n",sizeofp);
You cannot use the sizeof in this case, since p is a pointer, not an array, but since you allocate it, you already know:
main()
{
size_t arr_size = 2000;
char *p=NULL;
p=malloc(arr_size * sizeof(char));
printf("size of p = %d\n",arr_size);
}
Edit - If the malloc fails to allocate the size you wanted, it won't give you a pointer to a smaller buffer, but it will return NULL.