Freeing a pointer in a structure referenced by a pointer - c

I have a pointer to several structures that have been allocated memory via:
STRUCTNAME *ptr;
ptr = (STRUCTNAME *)malloc(sizeof(STRUCTNAME)*numberOfStructs);
The structures are accessed via a offset like so:
(ptr + i)->field;
The structures have 2 fields that are character pointers as follows:
typedef struct
{
char *first;
char *second;
}STUCTNAME;
These fields are allocated memory as follows:
(ptr + i)->first = (char *)malloc(strlen(buffer));
This appears to work but when I try to free the pointers within the structures I get a segmentation fault 11 when I do this:
free((prt + i)->first);
Help?
Notes:
buffer is a character array. Offsetting a pointer by a integer should increment the pointer by the size of what it is pointing to times the integer correct?
Here is a link to my full source code. I have not written some of the functions and I am not using the freeAllpointers and printAll yet.
https://drive.google.com/file/d/0B6UPDg-HHAHfdjhUSU95aEVBb0U/edit?usp=sharing
OH! Thanks everyone! Have a happy Thanksgiving! =D (If you're into that kinda stuff)

In case, you don't initialize all those members in that piece of code, you're not showing us:
Allocate the struct storage (STRUCTNAME*) with calloc(), so that all allocated memory, namely firstand second are zero at the beginning. Passing NULL to free() will result in a no-op. Passing any wild (garbage) pointer to free() may cause a segmentation fault.
To detect a double-free, set ptr[i].first = NULL; after free(ptr[i].first); as a defensive measure for testing.
Notes: buffer is a character array. Offsetting a pointer by a integer
should increment the pointer by the size of what it is pointing to
times the integer correct?
Yes, except for void* on those compilers, which don't define sizeof(void), which is defined to have undefined behavior, to a value > 0: What is the size of void?
Edit:
void makeReviews(FILE *input, REVIEW *rPtr, int numReviews) <-- This does NOT return the new value of rPtr. In main(), it will remain NULL.
Do something like this:
REVIEW* makeReviews(FILE *input, int numReviews);
//...
int main(){
//...
rPtr = makeReviews(input,numReviews);
//...
}
or
void makeReviews(FILE** input,REVIEW** rPtrPtr,int numReviews){
REVIEW* rPtr = *rPtrPtr;
//...
*rPtrPtr = rPtr;
}
//...
int main(){
//...
makeReviews(input,&rPtr,numReviews);
//...
}
fgets(cNumReviews, sizeof(cNumReviews), input); <-- Perhaps, you could use something like fscanf().

Related

Different arrays in dynamic struct, creation, writting, reading

Strait to the point.
I have a struct with a string, char and int.
The struct is created dynamically because i will need it in different parts of my program.
struct A
{
char staticString[20];
char* dynamicString;
char character;
int integer;
};
I know if i want to create a struct i call:
A example = (A)malloc(sizeof(A));
In order to populate the dynamicString and int i used:
example->dynamicString = (char*)malloc(sizeof(char*));
example->integer = (int)malloc(sizeof(int));
Unfourtanetly when i tried to populate staticString and char it didn't worked.
Don't even ask what was my code for those, i tried a lot of combinations from everywhere.
In addition to that can somebody show me examples how to write/read those values?
Thanks in advance.
First things first:
You're using C, and by the way you've defined the structure, you need to declare the pointer like so:
struct A *example;
Next, malloc returns a pointer, so you need to cast to a pointer (and not to a structure):
(struct A *)malloc(sizeof(struct A));
Secondly, I'm not sure why but hey:
- you're trying to dynamically allocate an int in the structure. As I said previously, malloc returns a pointer, so in your structure you need an int pointer like so "int *integer;"
- you're trying to allocate a dynamic string, however you're not doing it properly, here is what I think you want
example->dynamicString = (char *)malloc(sizeof(char) * 10);
Where 10 is the size of your dynamic string.
Edit:
you may also populate the integer in your struct statically or dynamically, but I think you intended the static approach:
example->integer = 123;
The dynamic approach would be (assuming you have int *integer in your struct):
example->integer = (int *)malloc(sizeof(int));
*(example->integer) = 123;
Every time you create a new struct the memory in the heap is set to size of :
sizeof(char)*20 + sizeof(char pointer) +sizeof(char)+ sizeof(int).
If you want to save a string that will be pointed to by your char pointer- then you ask for allocation in heap for the size of that string- and malloc returns the pointer to that memory allocation on heap.
So, you already have a space for your char array, char pointer, char and int that was allocated when you asked to make a new struct and do not need to allocate it again.
also, keep in mind malloc returns a pointer to the allocated place on the heap- so if you malloc(sizeof(int)) you get a pointer to a memory allocation for an int on the heap- which is pointed to by a int pointer Not an int.
good luck!

C - Storing pointers with malloc() in an array, can't free() them afterwards

I want to store pointers that have been allocated using malloc() in an array and then free all of them after. However even though the program doesn't complain it doesn't work. Below cleanMemManager() won't actually free the memory as when tested inside main() the char* pointer is not NULL and it will print ???.
code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void **ptrList = NULL;
void tfree(void** ptr)
{
free(*ptr);
*ptr = NULL;
}
void* talloc(int size)
{
void* ptr = malloc(size);
ptrList[0] = ptr; ///No clue if this actually does what I think it does
return ptrList[0];
}
void initMemManager()
{
ptrList = (void**)malloc(sizeof(void**) * 3);
memset(ptrList, 0, sizeof(void**) * 3);
}
void cleanMemManager()
{
tfree(&ptrList[0]); //Doesn't free the right pointer it seems
}
int main()
{
initMemManager();
char* ptr = (char*)talloc(3);
cleanMemManager();
if (ptr != NULL) //This will trigger and I'm not expecting it to
printf("???");
getchar();
return 0;
}
I don't understand the syntax to use for this, does the pointer not actually get touched at all? What is it freeing then since it doesn't throw any errors?
In main, char *ptr = (char*)talloc(3); declares ptr to be a local variable. It contains a copy of the value returned by talloc, and none of your subroutines know about ptr or where it is. So none of them change the value of ptr. Thus, when you reach if (ptr != NULL), the value of ptr has not changed.
Additionally:
In initMemManager, you should use sizeof(void *) in two places where you have sizeof(void**). In these places, you are allocating and copying void * objects, not void ** objects.
It looks like you are trying to implement a sort of smart memory manager that automatically sets pointers to NULL when they are freed. To do that in C, you would have to give up having copies of pointers. For example, ptr is a copy of ptrList[0], but tfree only sets whichever copy it is passed to NULL. We could give advice on building such a system, but it would quickly become cumbersome—your memory manager needs to keep a database of pointers and their copies (and pointers derived from them, as by doing array arithmetic). Or you have to refer to everything indirectly through that ptrList array, which adds some mess to your source code. Unfortunately, C is not a good language for this.
Freeing doesn't guarantee that pointers pointing to the allocated block will be set to NULL. If you actually try doing
if (ptrList[0] != NULL)
printf("ptrList[0] != NULL");
you will see that the program won't output and if you remove the cleanMemManager() function call, it will output. This means tfree function is working as intended, it's freeing the memory that was allocated.
Now as to why ptr variable being not set to NULL, it's simply because ptr is still storing the old address. cleanMemManager() has no way of mutating the variable ptr. This is commonly called dangling pointer or use after free.
Also free() doesn't clean/zero out the the allocated space, the block is simply marked as "free". The data will most likely remain in the memory for a moment until the free block is overwritten by another malloc request.

Why does this example of pointer dereferencing work?

I have some code, and it works, and I don't understand why. Here:
// This structure keeps the array and its bookkeeping details together.
typedef struct {
void** headOfArray;
size_t numberUsed;
size_t currentSize;
} GrowingArray;
// This function malloc()'s an empty array and returns a struct containing it and its bookkeeping details.
GrowingArray createGrowingArray(int startingSize) { ... }
// Self-explanatory
void appendToGrowingArray(GrowingArray* growingArray, void* itemToAppend) { ... }
// This function realloc()'s an array, causing it to double in size.
void growGrowingArray(GrowingArray* arrayToGrow) { ... }
int main(int argc, char* argv[]) {
GrowingArray testArray = createGrowingArray(5);
int* testInteger = (int*) malloc(1);
*testInteger = 4;
int* anotherInteger = (int*) malloc(1);
*anotherInteger = 6;
appendToGrowingArray(&testArray, &testInteger);
appendToGrowingArray(&testArray, &anotherInteger);
printf("%llx\n", **(int**)(testArray.headOfArray[1]));
return 0;
}
So far, everything works exactly as I intend. The part that confuses me is this line:
printf("%llx\n", **(int**)(testArray.headOfArray[1]));
By my understanding, the second argument to printf() doesn't make sense. I got to mostly by trial and error. It reads to me as though I'm saying that the second element of the array of pointers in the struct is a pointer to a pointer to an int. It's not. It's just a pointer to an int.
What does make sense to me is this:
*(int*)(testArray.headOfArray[1])
It's my understanding that the second element of the array of pointers contained in the struct will be fetched by the last parenthetical, and that I then cast it as a pointer to an integer and then dereference that pointer.
What's wrong with my understanding? How is the compiler interpreting this?
My best guess is that your appendToGrowingArray looks something like this:
void appendToGrowingArray(GrowingArray* growingArray, void* itemToAppend) {
growingArray->headOfArray[growingArray->numberUsed++] = itemToAppend;
}
though obviously with additional logic to actually grow the arrow. However the point is that the itemToAppend is stored in the array pointed to by headOfArray.
But, if you look at your appendToGrowingArray calls, you are passing the addresses of testInteger and anotherInteger -- these are already pointers to integers, so you are storing pointers to pointers to integers in your headOfArray when you really intend to store pointers to integers.
So, when you consider testArray.headOfArray[1], it's value is the address on main's stack of the variable anotherInteger. When you dereference it the first time, it now points to the address of the buffer returned by the second malloc call that you stored in anotherInteger. So, it's only when you deference it a second time that you get to the contents of that buffer, namely the number 6.
You probably want to write:
appendToGrowingArray(&testArray, testInteger);
appendToGrowingArray(&testArray, anotherInteger);
instead.
(As noted in a comment, you also should fix your mallocs; you need more than 1 byte to store an integer these days!)

I am trying to allocate dynamic memory to element of structure and getting segmentation fault

I am trying to allocate dynamic memory to element of array but i am getting segmentation fault
This is the code :
#include <stdio.h>
#include <stdlib.h>
struct Octstr
{
unsigned char *data;
long len;
long size;
int immutable;
};
int main()
{
struct Octstr *obj;
obj->data = (char*)malloc(16);
return 0;
}
You didn't allocate memory for the struct itself: obj is uninitialised.
The behaviour of your code is undefined.
Either malloc some memory for it or use automatic storage duration:
struct Octstr obj;
obj.data = (char*)malloc(16);
Finally, don't forget to balance your malloc calls with calls to free. (Also it's unnecessary to cast the return pointer of malloc in C, and do check the return pointer value to check if the allocation was successful).
struct Octstr *obj; is an uninitialized variable, it stores an unspecified address, and therefore points to an unspecified location.
Dereferencing it by trying to access a member is undefined behavior, which in your case results in a segmentation fault.
Initialize all your pointers by NULL.And make sure you free it after usage of your pointer at the end of your code.

memcpy-ing to a feild of a struct pointer in c

I am having trouble with struct pointers....Here are two examples in my code that are essentially doing the same thing except dsp is not a pointer and InMemory[Idx] is a pointer, how to I use memcpy in the pointer case?
my_struct* InMemory[SIZE]
//works prints: tmp3:local_file (file name)
memcpy(dsp.result.list[i].owner_name,req.file_name,256);
char tmp3[256];
memcpy(tmp3,dsp.result.list[i].owner_name,256);
printf("tmp3:%s\n",tmp3);
//doesn't work, prints: tmp:_____<---nothing! ??
//I am trying to copy the result from above into a field of the struct pointer array
char tmp2[256];
memcpy(InMemory[Idx]->filename,dsp.result.list[i].owner_name,256);
memcpy(tmp2,InMemory[Idx]->filename,256);
printf("tmp:%s\n",tmp2);
From your code, you have not allocated member elementes of InMemory
for (i=0;i<SIZE;i++)
{
// allocate elements here
InMemory[i]->filename = malloc(....)
// other allocations
}
// now use memcpy

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