I have a struct in my main function. I pass that pointer to another function which does some stuff and if conditions are met, it passes it to another function to get filled out. When returning to the main function the t struct contains none of the data mydata that was copied into it.
typedef struct _t {
int one;
int two;
int three;
int four;
} T;
void second(T *t) {
t = malloc(20);
memcpy(t, mydata, 20);
}
void first(T *t) {
second(t);
}
int main() {
T t;
first(t);
}
Do I need to be working with double pointers here? If the address of t was 0x1000 and I passed it to first() then wouldn't referencing t just be 0x1000? And same as if I pass the pointer to second()?
In this answer, I assume that, for reasons not shown, you do in fact need to make a dynamic memory allocation. If that is not the case, the only changes that need to be made are replacing first(t); with first(&t);, and removing t = malloc(20);.
The first problem to fix is that t in main should have the type T *, not T. You are making a dynamic memory allocation, and seem to want to store that pointer in t, so you would need: T *t;.
The second problem is that you want to manipulate the value of t in main, but are passing it by value to first. Instead, you need to pass a pointer to t into first: first(&t);.
Fixing both of these, you now pass a pointer to a pointer to T (the type of &t) into first and second, so you need to change their signatures to be, respectively, void first(T **t) and void second(T **t).
Applying both changes, as well as making some small style tweaks, we get:
typedef struct T {
int one;
int two;
int three;
int four;
} T;
void second(T **t_ptr) {
*t_ptr = malloc(20);
memcpy(*t_ptr, mydata, 20);
}
void first(T **t_ptr) {
second(t_ptr);
}
int main() {
T *t;
first(&t);
}
Another thing that's missing, and needs to be added, is checking for the success of malloc, but I haven't added that to the above code.
Also, what you've shown in the question shouldn't compile; you're passing a struct to a function that accepts a pointer.
Your problems are common to new C developers. And actually you have two of them.
The first problem is that you pass your structure by value.
The first function is declared to receive a pointer to T but you pass t and not &t (which is the address of t - and this is what you want when a function accepts a pointer).
However there is still another problem so that even if you change your code as suggested above it will still not work correctly. second allocates memory using malloc. The function receives T as a pointer T *t. You assign the output of malloc to t in effect overwriting what t points to (and if t was previously allocated you will leak memory here).
Bellow you can see a correct code for what you want.
typedef struct _t {
int one;
int two;
int three;
int four;
} T;
/* Make sure we have some data to initialize */
T mydata = {0};
/*
We take a pointer to a pointer and change what the external pointer points to. */
In our example when this function is called *ppt is NULL
and t is a pointer to t in main()
*/
void second(T **ppt) {
/*
We never calculate the size of structures by hand. It can change depending on
OS and architecture. Best let the compiler do the work.
*/
*ppt = (T*)malloc(sizeof(T));
memcpy(*ppt, &mydata, sizeof(T));
}
void first(T **ppt) {
/* Make sure we don't leave dangling pointers. */
if (NULL != *ppt)
free(*ppt);
second(ppt);
}
int main() {
T *t = NULL; /* A pointer to our data */
/*
We pass a pointer to our pointer so that the function can change the value it
holds
*/
first(&t);
/* Always do an explicit return if the type of the function is not void */
return 0;
}
How to understand what is going on:
First we declare t as a pointer to a memory holding a type T and we make sure we initialize the pointer to point to NULL (which is a convention meaning that the pointer is not initialized).
We have a function that will allocate the memory for us using malloc. malloc allocates memory from the heap and returns the address of that memory. (In reality a pointer is just a variable holding an address in memory). We want to place that address in t declared in main(). To do so we need to pass to the allocating function the address of t so it can be modified. To do this we use the address of operator - &. This is why we call the function like this first(&t).
Our allocating function accepts a pointer to a pointer. This is because we want to change the address t points to. So we declared the parameter as T **ppt. It holds the address of the pointer *t in main. In the function we dereference the pointer to the pointer to get the original pointer we want to assign the address malloc returns.
Related
If I have a struct like:
struct Cell
{
unsigned int *x, *y;
char flag;
};
Would the following constructor and deconstructors be sufficient for safely allocating and de-allocating memory?
// Constructor function.
struct Cell *Cell_new()
{
struct Cell *born = malloc(sizeof(struct Cell));
if (born == NULL)
return NULL;
born->x = NULL;
born->y = NULL;
born->flag = false;
return born;
}
// Deconstructor function.
// When called, use Cell_destroy(&cell);
char Cell_destroy(struct Cell **cell)
{
free(cell);
cell = NULL;
}
Is this correct?
One thing I don't understand is if I do:
struct Cell *myCell = Cell_new();
Cell_destroy(&myCell);
When I'm calling destroy, it is expecting an address to a pointer (a pointer to a pointer) and yet I'm providing an address to a struct.
What I mean is
My function expects:
Pointer -> Pointer -> Tangible Object
What I'm giving it:
Pointer -> Tangible Object
Is this flawed logic? I've been looking up this for awhile so it's safe to say I might be confusing myself.
It is correct to pass a parameter of type struct Cell ** to the destructor function because, this function will change the value of the pointer intended.
So, if you want to free the object and set the pointer to NULL, you can write:
void Cell_destroy(struct Cell **cell)
{
free(*cell);
*cell = NULL;
}
Note how cell is de-referenced.
I would consider the memory pointed to by the x and y pointers in the struct. If those pointers are non-null, what part of the code is responsible ("owns") that memory? A typical thing to do would be to check for null in Cell_destroy and if x or y is non-null, call free on them as well if it is the case that some other part of your Cell ADT (abstract data type) is allocating that memory dynamically via malloc. To be sure, free(cell) will only free memory used by the struct itself, not any memory pointed to by x and y.
Regarding Cell_destroy taking a pointer-to-a-pointer so it can set the passed-in pointer to null, that is an atypical way for a destructor function to work ... the way free itself works is more typical - nulling out the passed-in pointer is typically a job left for the caller of free.
Codes needs to free the same pointer it allocated. Free the de-referenced value.
Code is also missing a return value. Suggest simply using a void function.
Code should also free the field's pointers x and y.
Code should be tolerant of repeated calls with same pointer and a null pointer. Add if (*cell)
void Cell_destroy(struct Cell **cell) {
if (*cell) {
free((*cell)->x);
free((*cell)->y);
free(*cell);
*cell = NULL;
}
}
No, it is not correct.
The destroy function should not return char. It should have return type void.
Further the function should take a pointer instead of a double pointer and you should not set cell to null. And do not take address of myCell when calling destroy.
I have a problem with adding data to my tree using this function.
I am using codeblocks and when I run my program it gives me windows error box
#include <stdio.h>
#include <stdlib.h>
#include <stddef.h>
struct arb{
int data;
struct arb*FG;
struct arb*FD;
};
void remplit(struct arb*r,int i)
{
if (r==NULL)
{
r=malloc(sizeof(struct arb));
r->data=i;
r->FD=NULL;
r->FG=NULL;
}
else
{
if (i>(r->data))
{
remplit(r->FD,i);
}
else
{
remplit(r->FG,i);
}
}
}
struct arb * test=NULL;
int main()
{
remplit(test,5);
printf("%d",test->data);
return 0;
}
You are passing your pointer by value, not name.
remplit(test,5);
This sends the value of test to remplit.
r=malloc(sizeof(struct arb));
This points r, a local variable, to the allocated memory. This doesn't effect the value of test in main.
printf("%d",test->data);
test is still NULL, your attempt to de-reference it causes a seg fault.
You have a global pointer set to NULL. You then pass that pointer by value to another function that allocates dynamic memory to it with malloc.
The problem is that because the pointer is passed by value, the value of the global is unchanged (is still NULL) because the copy of your pointer now stores the address of the memory region allocated by malloc and not your global pointer.
Now when you dereference the pointer (after the call to remplit) it is giving you a segfault because test is still NULL.
If you want to use a function to allocate memory to a pointer that you pass to it, you need to make the function take a double pointer and assign the return value of malloc to the dereference of the pointer.
As a simple example, consider the following for creating a char array using double pointer and a utility function that does the allocation
void create_char_array(char** p, int size)
{
*p = NULL;
*p = malloc(size * (sizeof char));
/* *p now points to dynamically allocated memory */
}
int main()
{
char* my_array;
/* allocate memory for an array of 10 chars using above function
by passing the address of my_array */
create_char_array(&my_array,10);
if (my_array != NULL)
{
/* you can now safely assign values to valid the indices in the array */
}
/* release memory */
free (my_array);
return 0;
}
If you learn to use a debugger, you can step through your code at all steps. You should see that inside your function the memory is allocated to your r pointer (which is a copy of test), but that the value of the global test pointer is still NULL. Your code actually has a memory leak because the copy of the pointer inside your function is destroyed when the function exits. Memory has been allocated to it but there is no way to free it as the variable no longer exists.
I learned from this page: FAQ that, if you want to initialize a pointer inside a function, then you should pass a pointer to pointer, i.e, **p as foo1()
void foo1(int **p) {
*p = malloc(100*sizeof(int)); // caller can get the memory
}
void foo2(int *p) {
p = malloc(100*sizeof(int)); // caller cannot get the memory
}
However, a pointer means its value is the address it points to. Where does the memory allocated in foo2() go after leaving its scope?
I still can't figure out the different behavior between passing pointer to value and pointer to pointer? I search through SO but only found the solution or short description. Could anyone help in more detail?
The memory allocated in foo2 is lost. This creates a memory leak because you have no idea where to find and use the allocated memory after foo2 returns.
Consider:
int *mymemory = NULL;
foo2(mymemory);
//mymemory is still NULL here. Memory has been allocated,
//but you don't know at which address
//in particular, you will never be able to free() it
versus:
int *mymemory = NULL;
foo1(&mymemory);
//mymemory is now the address of the memory
//allocated by the function
dostuffwith(mymemory);
free(mymemory);
Maybe it helps if we start with only one level of indirection.
consider this:
void foo1(int *p) {
^^
//this p is local to the foo1 function
//p contains the address of an int
*p = 12;
//now we dereference the pointer, so we set what p points to , to 12
}
void func(void) {
int x;
^^
//here is the x
foo1(&x);
^^
//now we find the location (address of) x, we copy that address
//into the arguments for foo1()
//foo1 sets our x int to 12
}
Let's add one more indiretion:
void foo1(int **p) {
^^
//this p is local to the foo1 function
//p contains the address of a pointer to an int
*p = NULL;
//now we dereferenced the pointer, so we get an int*. We just
//set it to NULL
}
void func(void) {
int *x;
^^
//here is the x.
foo1(&x);
^^
///now we find the location (address of) x, we copy that address
//into the arguments for foo1()
//foo1() sets the x pointer to NULL.
}
In both cases we are able to manipulate the x variable inside func1(), since the location(address of)
the x variable is passed into func1().
In the last case, we did *p = NULL;. Which would make x == NULL. We could have set it
to something that malloc() returned: *p = malloc(100)
But if we alter the first case:
void foo1(int *p) {
^^
//this p is local to the foo1 function
//p contains the address of an int
p = NULL;
//now we just set the local `p` variable to NULL.
//the caller will not see that, since `p` is just our own copy
//of pointer.
}
void func(void) {
int x;
^^
//here is the x
foo1(&x);
//foo1 just set its own copy of the pointer we created by doing `&x` to NULL.
//we will not see any changes here
}
We just set p = NULL; in the last case here. If we used malloc instead:
void foo1(int *p) {
^^
p = malloc(100);
//now we just set the local `p` variable to what malloc returns.
//the caller will not see that, since `p` is just our own local copy
//of the pointer.
//When foo1() returns, noone has any way of knowing the location
//of the memory buffer that malloc returned, so this memory is lost (a memory leak)
}
In your second example the memory allocated is leaked - once foo2 ends, there is no variable left that contains the address that was allocated, so it can't be freed.
You could also consider
void foo3 (int bar) {
bar = 8;
}
int main (int argc, char *argv[]) {
int x = 0;
foo3(x);
printf("%d\n", x);
return 0;
}
when foo3 ends, x is still 0 - the change to the contents of bar in foo3 don't affect the outer variable that was passed in. You're doing exactly the same when you pass in a single pointer - you're assigning the address of some memory to it, but then losing that address when the function exits.
To better understand indirection levels in C, it can be instructive to look at how the compiler organizes its memory.
Consider the following example :
void function1 (int var1, int var2) { ... }
In this case, function1 will receive 2 variables. But how ?
These variables will be put into the call stack memory. This is a linear, LIFO (Last in, First Out) type of allocation strategy.
Before calling function1(), the compiler will put var1 then var2 into the call stack, and increment the position of the call stack ceil. Then it will call function1(). function1() knows it must get 2 arguments, and so it finds them into the call stack.
What happens after function1() finishes ? Well, the call stack is decremented, and all variables into it are simply "disregarded", which is almost the same as "being erased".
So it's pretty clear that whatever you do to these variables during function1() is going to be lost for the calling program. If anything has to remain available to the calling program, it needs to be provided into a memory space that will survive the call stack decrement step.
Note that the logic is the same for any variable inside function1() : it will be unavailable to the calling function after function1() finishes. In essence, any result still stored into function1() memory space is "lost".
There are 2 ways to retrieve a usable result from a function.
The main one is to save the result of the function into a variable of the calling program/function. Consider this example :
int* foo3(size_t n) { return (int*) malloc(n); }
void callerFunction()
{
int* p;
p = foo3(100); // p is still available after foo3 exits
}
The second, more complex, one is to provide as an argument a pointer to a structure which exists into the calling memory space.
Consider this example :
typedef struct { int* p; } myStruct;
void foo4(myStruct* s) { s->p = (int*) malloc(100); }
void callerFunction()
{
myStruct s;
foo4(&s); //p is now available, inside s
}
It is more complex to read, but also more powerful. In this example, myStruct contains a single pointer, but the structure could be a lot more complex. This open the perspective to offer myriad of variables as the result of a function, instead of being limited to basic types, as for the previous example with foo3().
So what happens when you know that your structure is in fact a simple basic type ? Well, you can just provide a pointer to it. And, by the way, pointer is itself a basic type. So if you want to get the result of a modified pointer, you can provide as an argument, a pointer to a pointer. And there we find foo1().
void foo1(int **p) {
*p = (int *) malloc(100); // caller can get the memory
}
The problem with foo2 is that the p which is passed in is only modified inside the foo2 function. This is the same as :
void bar(int x)
{
x = 42;
}
...
int a = 7;
bar(a);
...
In the above code, a doesn't change because of the call to bar. Instead, a copy of a is passed to bar, and the copy is modified in bar.
The exact same thing happens in foo2. The memory is allocated, stored in p, which is a copy of the pointer passed in. When the code returns, the original pointer retains its original value.
By passing the address of a pointer (&ptr) to foo1, we can modify the ORIGINAL pointer, and thus pass the address of the allocation back to the caller of foo1.
Of course, when there is no reference back to the originally allocated memory, as is the case after a call to foo2, it is called a memory leak - generally considered a bad thing.
Passing a pointer to value: A copy of the pointer(i.e. address of the value) is made in the function(on the stack frame). This allows you to modify the value.
Passing a pointer to a pointer: A copy of the pointer to a pointer(i.e. address of the pointer which in turn points to the value) is made in the function(on the stack frame). This allows you to modify the value as well as the pointer to this value.
The memory allocated using malloc, calloc, realloc and new resided on the heap which means that it exists even after a function returns(stack frame destroyed).
void foo2(int *p) {
p = (int *) malloc(100); // caller cannot get the memory
}
However, since the pointer p is lost after the function is returned, this memory cannot be accessed and will result in a leak.
As behaviour of all arguments is the same as local variables (they are passed by value), you can not modify pointer passed by value.
So in foo2() you allocate memory, but you can not use it outside the function, as you actually modify local variable.
The foo() function actually modifies the value pointed by **p, so pointer passed to function will be updated.
OK, this is the definition of the struct:
typedef struct {
int first;
int last;
int count;
char * Array [50];
} queue;
and I use another function to initialize it
void initialize(queue * ptr){
ptr=malloc(sizeof(queue));
ptr->first=0;
ptr->last=0;
ptr->count=0;
}
Then I use printf to print out first, last and count. All three should be zero. However, what I actually get is, count is 0 as I expected, but first&last are two very large strange numbers and they change every time I run the program. Can anybody tell me what's wrong here? Thank you.
You are passing your pointer by value. The function changes a copy of the argument it receives, but the caller's pointer is not modified and is probably unintialized.
You need to change your function to take a queue** and pass the address of the pointer you want to initialize.
Alternatively you could return a pointer instead of passing it in as an argument. This is a simpler approach.
Given:
void initialize(queue * ptr);
Pass it like this:
queue q; // caller allocates a queue
initialize(&q);
// now q is initialized
Also, it's allocated by the caller -- don't malloc it.
// bad
void initialize_bad(queue * ptr){
ptr=malloc(sizeof(queue)); << nope. already created by the caller. this is a leak
ptr->first=0;
ptr->last=0;
ptr->count=0;
}
// good
void initialize_good(queue * ptr){
ptr->first=0;
ptr->last=0;
ptr->count=0;
// ptr->Array= ???;
}
If you prefer to malloc it, then consider returning a new allocation by using this approach:
queue* NewQueue() {
// calloc actually works in this case:
queue* ptr = (queue*)calloc(1, sizeof(queue));
// init here
return ptr;
}
Ultimately, what is 'wrong' is that your implementation passes a pointer by value, immediately reassigns the pointer to a new malloc'ed allocation, initializes the malloc'ed region as desired, without ever modifying the argument, and introducing a leak.
Here is the smallest alteration to your program which should correct your problem:
void initialize(queue * * pptr) {
queue * ptr;
ptr=malloc(sizeof(queue));
if (ptr) {
ptr->first=0;
ptr->last=0;
ptr->count=0;
}
/* The assignment on the next line assigns the newly allocated pointer
into memory which the caller can access -- because the caller gave
us the address of (i.e. pointer to) such memory in the parameter
pptr. */
*pptr = ptr;
}
The most important change is to pass a queue ** to your initialize function -- otherwise you are changing a copy of the queue * supplied as the actual parameter when you call initialize(). By passing a pointer to the pointer, you can access the original variable which stores the pointer in your caller.
I couldn't resist and also added a check for NULL returned from malloc(). That doesn't address your problem, but I couldn't bring myself to post code that didn't do it.
I thought I understood the basics of pointers, but after checking out some documentation on some sqlite3 methods I got thrown, so now I am unsure if my understanding is correct.
Here is a call to an sqlite3 method:
char* dataFilePath = "foobar.sqlite";
if (sqlite3_open(dataFilePath, &database) != SQLITE_OK) {...}
And here is the function header declaration:
int sqlite3_open(
const char *filename, /* Database filename (UTF-8) */
sqlite3 **ppDb /* OUT: SQLite db handle */
);
Why is it that &database suddenly becomes a pointer to a pointer?
Another method call to close the database connection is:
sqlite3_close(database);
With the following at the function header:
int sqlite3_close(sqlite3 *);
Why is this just a pointer, when I pass in a pointer? Would this not be a pointer to a pointer?
From all examples I have seen it always seemed the inverse of the functions above, ie.
// function
void foo(someDataType *bar) { ... }
// function call
foo(&bar);
Thanks for the help.
Most likely, sqlite3_open is allocating memory for the database handle. For this reason the function needs a pointer to a pointer to the database handle (sqlite3) so that it can modify the pointer to the database handle. For example:
typedef struct { /*...*/ } sqlite3;
int sqlite3_open(const char *filename, sqlite3 **ppDb) {
/* ... */
// Allocate memory for the database handle.
*ppDb = (sqlite3 *)malloc(sizeof(sqlite3));
/* ... */
return 0;
}
However, sqlite3_close only needs a single pointer to free the memory:
int sqlite3_close(sqlite3 *pDb) {
/* ... Cleanup stuff ... */
free(pDb);
return 0;
}
I think the short explanation for what you're asking is that using "&" essentially means "a pointer to this"
int value = 0;
int *pointer = &value;
int **doublePointer = &pointer;
A pointer is the address of a variable.
Assuming that database is declared as sqlite3* database;, &database is the address of (or, a pointer to) the database pointer.
sqlite3_open takes a pointer to a pointer so that it can set the value that the pointer points to. It makes a sqlite value, and changes your pointer to point to it. sqlite3_close doesn't change what the pointer points to, so all it needs is the pointer itself.
As usual, the C FAQ List contains relevant information. See I have a function which accepts, and is supposed to initialize, a pointer: and Does C even have "pass by reference"?.
i don't know what you want to do with sqlite function. But using pointers makes you to keep changes in functions.
When you pass a variable to a function, the variable will be duplicated.
for example
int var1=0;
in *ptr1=&var1;
func(var1, ptr1);
the value of var1=5
the adress of var1 = 0xff2200 (something like that)
the value of ptr1 = 0xff2200 (the adress of var1)
the adress of ptr1 = 0xff0022 (something different)
Lets write a function which uses these two var as arg
void func1(int x, int *p){
x+=5;
(*p)-=5;
}
after u use this function;
func(var1, ptr1);
var1 will not equal to 0!!! İt will be -5
Because;
in function func1
the value of x = 0 (the value of var1)
the adress of x = 0xaabbcc (something different then var1!!! this is why x+=5 is not effective on var1. It happens in another part of memory! When u return, this memory will be free again. And you'll lose your changes...)
the adress of p = 0xcccccc (something different too)
the value of p = 0xff2200 (the value of ptr1 and the adress of var1! This operation will be done in the var1's adress so you will not lose your changes)
İf we have to keep our changes of variables -in functions-, we have to use pointers for those var.
İf our variable keep an adress, it means; it is a pointer. And if we want to keep changes of pointer -in functions- then we have to use pointer to pointer.
This is my first message and i hope this will be helpfull...
And "pass by reference" means "pass by pointer" other languages don't use pointers. so you have to pass by reference sometimes. But in C, pointers will do its job...