Assume I have a struct node like such:
struct node {
struct other_struct value;
int some_num;
}
I have seen snippets of code where the struct can be initialized without calling malloc, like this:
struct node my_node;
my_node.value = NULL;
my_node.some_num = 2;
And then value can later be malloced. However, how would I free my_node?
my_node is allocated on the stack, and once it goes out of scope, its memory is automatically deallocated.
Use malloc if you want to allocate something on the heap, and it will persist until you free it. For your example, you would do:
struct node *my_node = malloc(sizeof(struct node));
my_node->some_num = 2;
// Sometime later
free(my_node);
Space will also be allocated for the value field if you malloc (as long as you pass the right size) or if you declare on the stack. sizeof(struct node) includes the size of other_struct.
Currently with those code snippets you can't really free value or my_node, as they're both just static variables, if in your node structure, the other_struct field was a pointer, then you could dynamically allocate some memory, and save the address in that pointer.
Then when done, free the pointer value.
If then, my_node was also a pointer (which I think is what you want), you would need to allocate memory for the node struct, and save the address to the pointer my_node. THEN, allocate some memory for the other_struct struct and save it to the pointer value. And after you're done, you would free value FIRST, then free my_node.
When doing things like this, I generally create a little constructor/destructor function, which will do it all for me. It can be too easy to forget to free the inner pointer (value) and just free the outer pointer (mynode). Then that would cause a memory-leak, as value would still be allocated and taking up room in memory.
The code:
struct node {
struct other_struct value;
int some_num;
};
defines value to be a member of struct node. It is a part of struct node. Whenever a struct node is created, it will contain the member named value, and no other memory needs to be allocated for value. Whether a struct node is created automatically or by malloc or other means, its memory will always include memory for value.
If you changed it to:
struct node {
struct other_struct *value;
int some_num;
}
then value would be a pointer to a struct other_struct, and you would need to provide memory for it to point to.
In this code:
struct node my_node;
my_node.value = NULL;
my_node.some_num = 2;
the object my_node is created automatically, and you do not have to do anything to release it or its members; that will happen automatically. You cannot use my_node.value = NULL because value is a structure, not a pointer.
If the structure definition is changed so that value is a pointer, then you can set to NULL, or you can set it to point at an existing object, or you can allocate memory for it and set it to point to the allocated memory. If you allocate memory for it, then you should ensure that memory is later freed (except it is okay not to free it if you are intentionally keeping it to the end of program execution anyway, and you are executing in user mode on a general-purpose operating system).
You don't free it... it is destroyed, when the block it has been declared in is abandoned by the program.
Let's say that this variable is defined at the beginning of a function's body block. When the function is being called, the function creates this variable as it's a local. When the function terminates, all local variables are destroyed, so it is your local structure.
If you declare such a variable out of any function, then your variable has global duration, and it is created at program initialization, and lives as much as the program itself.
Related
Could someone please explain to me the difference between creating a structure with and without malloc. When should malloc be used and when should the regular initialization be used?
For example:
struct person {
char* name;
};
struct person p = {.name="apple"};
struct person* p_tr = malloc(sizeof(struct person));
p_tr->name = "apple";
What is really the difference between the two? When would one approach be used over others?
Having a data structure like;
struct myStruct {
int a;
char *b;
};
struct myStruct p; // alternative 1
struct myStruct *q = malloc(sizeof(struct myStruct)); // alternative 2
Alternative 1: Allocates a myStruct width of memory space on stack and hands back to you the memory address of the struct (i.e., &p gives you the first byte address of the struct). If it is declared in a function, its life ends when the function exits (i.e. if function gets out of the scope, you can't reach it).
Alternative 2: Allocates a myStruct width of memory space on heap and a pointer width of memory space of type (struct myStruct*) on stack. The pointer value on the stack gets assigned the value of the memory address of the struct (which is on the heap) and this pointer address (not the actual structs address) is handed back to you. It's life time never ends until you use free(q).
In the latter case, say, myStruct sits on memory address 0xabcd0000 and q sits on memory address 0xdddd0000; then, the pointer value on memory address 0xdddd0000 is assigned as 0xabcd0000 and this is returned back to you.
printf("%p\n", &p); // will print "0xabcd0000" (the address of struct)
printf("%p\n", q); // will print "0xabcd0000" (the address of struct)
printf("%p\n", &q); // will print "0xdddd0000" (the address of pointer)
Addressing the second part of your; when to use which:
If this struct is in a function and you need to use it after the function exits, you need to malloc it. You can use the value of the struct by returning the pointer, like: return q;.
If this struct is temporary and you do not need its value after, you do not need to malloc memory.
Usage with an example:
struct myStruct {
int a;
char *b;
};
struct myStruct *foo() {
struct myStruct p;
p.a = 5;
return &p; // after this point, it's out of scope; possible warning
}
struct myStruct *bar() {
struct myStruct *q = malloc(sizeof(struct myStruct));
q->a = 5;
return q;
}
int main() {
struct myStruct *pMain = foo();
// memory is allocated in foo. p.a was assigned as '5'.
// a memory address is returned.
// but be careful!!!
// memory is susceptible to be overwritten.
// it is out of your control.
struct myStruct *qMain = bar();
// memory is allocated in bar. q->a was assigned as '5'.
// a memory address is returned.
// memory is *not* susceptible to be overwritten
// until you use 'free(qMain);'
}
If we assume both examples occur inside a function, then in:
struct person p = {.name="apple"};
the C implementation automatically allocates memory for p and releases it when execution of the function ends (or, if the statement is inside a block nested in the function, when execution of that block ends). This is useful when:
You are working with objects of modest size. (For big objects, using many kibibytes of memory, malloc may be better. The thresholds vary depending on circumstances.)
You are working with a small number of objects at one time.
In:
struct person* p_tr = malloc(sizeof(struct person));
p_tr->name = "apple";
the program explicitly requests memory for an object, and the program generally should release that memory with free when it is done with the object. This is useful when:
The object must be returned to the caller of the function. An automatic object, as used above, will cease to exist (in the C model of computation; the actual memory in your computer does not stop existing—rather it is merely no longer reserved for use for the object) when execution of the function ends, but this allocated object will continue to exist until the program frees it (or ends execution).
The object is very large. (Generally, C implementations provide more memory for allocation by malloc than they do for automatic objects.)
The program will create a variable number of such objects, depending on circumstances, such as creating linked lists, trees, or other structures from input whose size is not known before it is read.
Note that struct person p = {.name="apple"}; initializes the name member with "apple" and initializes all other members to zero. However, the code that uses malloc and assigns to p_tr->name does not initialize the other members.
If struct person p = {.name="apple"}; appears outside of a function, then it creates an object with static storage duration. It will exist for the duration of program execution.
Instead of struct person* p_tr = malloc(sizeof(struct person));, it is preferable to use struct person *p_tr = malloc(sizeof *p_tr);. With the former, a change to the p_tr requires edits in two places, which allows a human opportunity to make mistakes. With the latter, changing the type of p_tr in just one place will still result in the correct size being requested.
struct person p = {.name="apple"};
^This is Automatic allocation for a variable/instance of type person.
struct person* p_tr = malloc(sizeof(person));
^This is dynamic allocation for a variable/instance of type person.
Static memory allocation occurs at Compile Time.
Dynamic memory allocation means it allocates memory at runtime when the program executes that line of instruction
Judging by your comments, you are interested in when to use one or the other. Note that all types of allocation reserve a computer memory sufficient to fit the value of the variable in it. The size depends on the type of the variable. Statically allocated variables are pined to a place in the memory by the compiler. Automatically allocated variables are pinned to a place in stack by the same compiler. Dynamically allocated variables do not exist before the program starts and do not have any place in memory till they are allocated by 'malloc' or other functions.
All named variables are allocated statically or automatically. Dynamic variables are allocated by the program, but in order to be able to access them, one still needs a named variable, which is a pointer. A pointer is a variable which is big enough to keep an address of another variable. The latter could be allocated dynamically or statically or automatically.
The question is, what to do if your program does not know the number of objects it needs to use during the execution time. For example, what if you read some data from a file and create a dynamic struct, like a list or a tree in your program. You do not know exactly how many members of such a struct you would have. This is the main use for the dynamically allocated variables. You can create as many of them as needed and put all on the list. In the simplest case you only need one named variable which points to the beginning of the list to know about all of the objects on the list.
Another interesting use is when you return a complex struct from a function. If allocated automatically on the stack, it will cease to exist after returning from the function. Dynamically allocated data will be persistent till it is explicitly freed. So, using the dynamic allocation would help here.
There are other uses as well.
In your simple example there is no much difference between both cases. The second requires additional computer operations, call to the 'malloc' function to allocate the memory for your struct. Whether in the first case the memory for the struct is allocated in a static program region defined at the program start up time. Note that the pointer in the second case also allocated statically. It just keeps the address of the memory region for the struct.
Also, as a general rule, the dynamically allocated data should be eventually freed by the 'free' function. You cannot free the static data.
I have the following structure that I've declared:
typedef struct binTreeNode
{
void *data;
struct binTreeNode *left;
struct binTreeNode *right;
} myBinaryTreeNode;
In my main function, I'm trying to use a pointer to an instance of that structure called root.
myBinaryTreeNode *root = malloc(sizeof(myBinaryTreeNode));
printf("\nPlease insert root data: ");
int input;
scanf("%d", &input);
root->data = (void*)&input;
printInt(root->data);
free(root);
this code runs perfectly well. But, I thought that when you have a struct with members that are pointers, you should free() each one of them (additionally to the pointer to the struct).
So here, I didn't malloc for root->data (because I think that mallocing the struct does that), but it is initialized to the input value and it's value gets printed successfully. When I try to free(root->data) my program crashes.
So root->data is not malloced when I malloc root? If not, how can I still use it?
Why does this happen? What am I missing here?
First, get the concept of where and how you need to (or need not) call free().
You don't need to malloc() "for" root->data, that variable is already allocated while you allocated memory equal to the size of the structure. Now, next, you certainly need root->data to point to some valid memory, so that you can dereference the pointer to read from to write into it. You can do that by either of two ways
store an address which is valid (like in your case, supply the address of an already existing variable)
assign a pointer returned by another malloc() (success).
In case, you're storing the pointer returned by malloc(), yes you must free() the memory but in your case, root->data holds a pointer not returned by malloc(), so it does not need free()-in either.
Just to add and emphasis on the part related to free()-ing, you must not attempt to free() the memory which has not been previously allocated by a call to malloc() and family otherwise, it invokes undefined behavior. To quote, standard C11, chapter §7.22.3.3, (emphasis mine)
void free(void *ptr);
The free function causes the space pointed to by ptr to be deallocated, that is, made
available for further allocation. If ptr is a null pointer, no action occurs. Otherwise, if
the argument does not match a pointer earlier returned by a memory management
function, or if the space has been deallocated by a call to free or realloc, the
behavior is undefined.
data member points to input local variable, so you cannot free it.
You must free members if you dynamically allocate them, like
typedef struct binTreeNode
{
int *data;
struct binTreeNode *left;
struct binTreeNode *right;
} myBinaryTreeNode;
root->data = malloc(sizeof(int));
scanf("%d", root->data);
free(root->data);
When I try to free(root->data) my program crashes.
You cannot free memory that you have not dynamically allocated. If you want to free the elements, you should first allocate them this way :
typedef struct binTreeNode
{
int *data;
struct binTreeNode *left;
struct binTreeNode *right;
} myBinaryTreeNode;
root->data = malloc(sizeof(int));
if (root->data == NULL)
{
printf("Error allocating memory\n");
return;
}
scanf("%d", root->data);
free(root->data);
Note that you should check the result of malloc before continuing.
So root->data is not malloced when I malloc root? If not, how can I still use it?
Note also that space for root->data is allocated when you allocate memory for the struct, but it is not malloced. You need an extra malloc especially for root->data, as shown in the example above. The reason why you need malloc for root->data is to be able to dereference the pointer.
I have added some annotations to your code.
I hope this clears things up.
// I have added an enclosing block for the purpose of demonstration:
{
// Here you allocate memory for a tree node object with dynamic/allocated
// storage duration.
// That is: three pointers, that point to nowhere are created and will
// exist until the memory holding them is free'd.
myBinaryTreeNode *root = malloc(sizeof(myBinaryTreeNode));
printf("\nPlease insert root data: ");
// Here you create an object with automatic storage duration.
// That is, the object is automatically destroyed when the function
// returns or the block is closed: { ... }
int input;
// here you attempt to assign a value to the variable
// (you should check the return value: the input may fail)
scanf("%d", &input);
// Here you assign the address of an object to a pointer.
// Note that you are only allowed to use the pointer as long as the
// object it points to "lives".
root->data = (void*)&input;
// Here you probably print the value, or the pointer address, or something else
printInt(root->data);
free(n); // <-- Where and how did you allocate this?
// Here you free the memory holding the tree node
free(root);
} // <-- At this point the "input" variable is automatically destroyed.
You should really read up on the different storage durations (automatic and dynamic/allocated) and the purpose of dynamic memory allocation:
http://en.cppreference.com/w/c/language/storage_duration
http://en.cppreference.com/w/c/language/lifetime
C Memory Management
To address the questions in detail:
When I try to free(root->data) my program crashes.
That is because you can only free() what you have malloc(). Memory management is different for the different types of storage duration.
So root->data is not malloced when I malloc root?
Memory for the pointer is allocated, but it doesn't "own" or "point to" any object by default.
If not, how can I still use it?
Just as you did. You assign a memory address of an object to it, that lives long enough for you to use it. You can produce such object using dynamic memory management (then take care that you destroy it manually once you're done: don't leak memory) or using "automatic memory management" as you have done (then take care that it isn't destroyed too early: don't access invalid memory). (To name just two of the possible ways to create an object...)
I have a structure:
struct generic_attribute{
int current_value;
int previous_value;
};
And a constructor which outputs a pointer to this structure:
struct generic_attribute* construct_generic_attribute(int current_value){
struct generic_attribute *ga_ptr;
ga_ptr = malloc (sizeof (struct generic_attribute));
ga_ptr->current_value = current_value;
ga_ptr->previous_value = 0;
return ga_ptr;
}
Now, in another function, I want to define a pointer and set it to point to the same address as the pointer that the above constructor outputs.
struct tagged_attribute* construct_tagged_attribute(int num_args, int *args){
...
struct generic_attribute* generic = malloc (sizeof(struct generic_attribute));
generic = construct_generic_attribute(args[0]);
...
}
It looks to me that what I am doing here is this:
1) I define a pointer "generic" and allocate a memory chunk to hold an instance of generic_attribute structure.
2) I call a function construct_generic_attribute within which, the program once again allocates a memory chunk of size of a generic_attribute structure. It outputs a pointer to this memory chunk.
3) In construct_tagged_attribute I set "generic" pointer equal to the pointer output by the construct_generic_attribute function, so now both of them point to the same memory slot.
However, it appears that I am allocating twice as much memory as I need to allocate.
Is there a way for me to allocate memory only once without getting a segmentation fault for failing to allocate space for "generic" pointer? Alternatively, am I misunderstanding what is happening in this code?
struct generic_attribute* generic = construct_generic_attribute(args[0]);
Should do the trick. Pointer variable is just that, a variable. You can trade pointer values around just like numbers.
Yes, you're misunderstanding, but I can't quite figure out what you think is happening to explain how it's wrong.
struct generic_attribute *generic = construct_generic_attribute(args[0]); a pointer is a kind of value. If you assign a pointer to another, you get two pointers to the same thing, without any allocation of memory. Since C doesn't manage memory for you, it's up to you to make sure that any object that's allocated is freed exactly once, and that you don't try to use pointers to an object after it's been freed.
Here
struct generic_attribute* generic = malloc (sizeof(struct generic_attribute));
you allocate a memory block, big enough to keep a generic_attribute structure, then store a pointer to that structure (technically: an address of the block) in the generic variable. Note: you do not initialize the structure members.
Then in
generic = construct_generic_attribute(args[0]);
you call a function, which internally allocates (another) block of memory and initializes it and returns a pointer to it (which was stored in a ga_ptr variable during the function execution). The pointer returned is then assigned to the generic variable, overwriting the value stored there by a previous instruction. Consequently you loose an access to the first allocated structure.
EDIT
I'm afraid I do not quite understand what you're trying to achieve. If you want two pointers to the same structure, just declarega1 and assign it a pointer to the created structure:
struct generic_attribute *ga1 = construct_generic_attribute(args[0]);
then make a copy of the pointer:
struct generic_attribute *ga2 = ga1;
I am a java programmer learning C. Have a question regaring functions. What are the differences between this:
main()
{
struct person myperson;
myperson = myfunction();
return;
}
struct person myfunction()
{
struct person myPerson;
myPerson.firstname = "John";
myPerson.lastname = "Doe";
return myPerson;
}
VS
main()
{
struct person *myperson;
myperson = myfunction();
return;
}
struct person* myfunction()
{
struct person *myPerson;
myPerson = malloc(sizeof(struct person));
myPerson->firstname = "John";
myPerson->lastname = "Doe";
return myPerson;
}
Are these legal in C? And y would 1 choose one over the other.
Thanks so much guys!
first code sample:
you create a struct in myfunction() on your stack and return it. then, you create another stack struct, and you copy the first to the second. the first is destroyed. the second will be automatically destroyed when you are out of the scope.
2 structs were actually created.
second code sample:
you create a struct in myfunction(), and then you copy only the address. the struct in main will actually be the same struct.
only one struct is created in here.
both code samples work, but for the later you will have to explicitly free the memory allocated for the struct, to avoid memory leak, but performance should be better since you don't need to copy the struct!
EDIT:
as mentioned by #Mat: this of course neglects the overhead of malloc(), which is not true for small structs.
The first version allocates the object on the stack and returns a copy of it. The second version creates the object on the heap and returns a pointer to it(this is closest to Java references except that the memory isn't automatically freed). You should not forget to call free() later on the returned pointer.
Btw, your main function is bad. It should be
int main(void)
{
...
return 0;
}
I suggest that you should read a good C book. This is really basic stuff you're asking.
I'm not sure if all this talk of "heap" and "stack" is cutting to the core of the language, so let me try something more language-intrinsic.
Your first version uses only automatic allocation, which means that all variables have automatic lifetime. That is, all variables end their life at the end of their enclosing scope: myFunction creates a local variable of type struct person and returns a copy of that variable; the main function declares a local variable of the same type and assigns to it the result of the function call. At the end of each scope, the local variables end as well.
The second version uses dynamic or manual allocation. You explicitly allocate storage for a person variable with the malloc() call, and that storage will remain allocated until someone deallocates is (via free()). Since you never deallocate it, this is in effect a memory leak.
The fundamental difference is one of lifetime and responsibility.
A few pros and cons: Automatic allocation means that responsibility is local, and you generally don't have to worry about anything. However, it comes at the price of having to copy arguments and return values by value, which may be expensive or undesirable. Manual allocation allows you to refer to large amounts of memory via a simple, cheap pointer, and is often the only way to implement certain constructions, but carries the burden of having the author remember who's responsible for which resource.
Both are legal, both work.
The 1st version is simpler, you avoid having to deal with memory allocation and releasing.
The 2nd version will perform better for bigger structs because you avoid putting the whole struct on stack for handing it over.
I would actually choose a third way. Let the caller worry about providing storage space (auto or dynamically allocated):
void myfunction(struct person* myPerson)
{
myPerson->firstname = "John";
myPerson->lastname = "Doe";
}
The function can be called either with an automatically or dynamically allocated variable:
struct person autoperson;
myfunction(&person);
struct person dynamic_person = malloc(sizeof struct person);
myfunction dynamic_person);
The first will allocate a struct person on the stack, and pass a copy of it back, then free the original. The second one will allocate it on the heap and pass a pointer to the location which was allocated, and will not free it.
The first one allocates the variables on the stack. The person object from myfunction is copied from the function and returned which is less efficient, but you can't get a memory leak which is good.
The second example returns a pointer (the *) to a person object that is dynamically allocated (with malloc). The person object allocated by malloc will never be destroyed unless you explicitly call free() on it, hwich you haven't - so you have a memory leak.
You need to explicitly free memory in C, it doesn't have garbage-collection like Java.
The first option creates a struct on the stack, when returning it, it gets copied to your struct defined in the main() function. Also copied are the fields. For larger structs this can be a costly operation.
The second option allocates dynamic memory, which does not get copied when you return it. You have to free() the pointer to avoid a memory leak.
Of course it depends on your needs, but for more important and long living objects I'd go for the second option. Also I would recommend to write allocation/initialization functions and a corresponding deallocation function. (see below why)
The problem is that the 2 strings you set in myfunction() are invalid outside of the function, as they are also created on the stack. You have to use strdup() or a similar function to make this failsave. Of course, to not let memory leaks slip in you have to free() the strduped pointers, just as with malloc().
In the first code, myPerson is an object of type struct person that is managed (*) by the implementation itself. In the second code, it is an object of type struct person * (a pointer to a struct person). In the second code, the object itself must be managed by the programmer (malloc, realloc, free).
Also, in the first code, the object itself is copied around a few times whereas in the 2nd code "only" the pointer gets copied. Usualy a pointer is much smaller than an object of a struct type.
Use the 2nd approach but remember to free the object.
Even better, create the object in the parent function and pass a pointer to functions: sruct person *myfunction(struct person *data) { /* ... */ }
(*) with object management I mean the time it gets created and deleted and stuff
First One:
main()
{
// create a person struct on the stack
struct person myperson;
// copy the struct returned by myfunction to myperson.
myperson = myfunction();
}
struct person myfunction()
{
// create a person struct on the stack.
struct person myPerson;
myPerson.firstname = "John";
myPerson.lastname = "Doe";
// return the myPerson struct. After myFunction returns, the memory
// holding the myPerson struct on the stack will be freed.
return myPerson;
}
Second one:
main()
{
// create a pointer to a person struct on the stack
struct person *myperson;
// assign the pointer returned by myfunction to myperson
myperson = myfunction();
}
struct person* myfunction()
{
// create a pointer to a person struct on the stack
struct person *myPerson;
// allocate memory for a person struct in dynamic memory and set myPerson
// to point to that memory. This memory will remain valid until it's freed by
// a call to the "free" function. Using malloc is much slower than creating
// an object on the stack. There is also the added performance cost of
// freeing the allocated memory at a later stage.
myPerson = malloc(sizeof(struct person));
myPerson->firstname = "John";
myPerson->lastname = "Doe";
// return the myPerson pointer
return myPerson;
}
I have a pointer to a struct. I call a routine that determines whether I need this struct and allocates space for it using malloc, returning the pointer to that area or zero if unused.
struct node *node_p;
node_p = need_this();
This works and I can properly access all the elements of the struct. One of the elements of struct node is *next which points to the next node struct and a pointer to a string but it's returning a pointer to a string that doesn't even exist in this routine.
node_p=find_string(node_p->next,"string");
However, this does return a pointer in the struct to the correct string.
struct node *node_p, *node_copy;
node_copy=find_string(node_p->next,"string");
The only difference is using a second pointer instead of the original. Am I doing something wrong here or must it be deeper into the called function? The problem with blaming the called function is I use it in multiple places for months without issue, however the other calls only look for the string and never go to 'next'.
EDIT: Further debugging has shown the problem actually lies with the pointer being passed to find_string and not the returned pointer. That changes the question and the problem so I need to open another question.
In this snippet:
struct node *node_p, *node_copy;
node_copy=find_string(node_p->next,"string");
you dereference node_p when it is not yet initialized (doesn't point to a legally allocated memory block). That's undefined behavior. You should set node_p to a legally allocated memory block of appropriate size first.
You need to allocate memory explicitly for every struct, i.e. allocate a new struct and set next pointer to point to it.
Pointers in C prior to initialization point to random place and you never should dereference them. Safe policy would be to init them to be NULLs.