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Dynamic memory access only works inside function
(1 answer)
Closed 4 years ago.
I have created a structure which includes an array of strings and have populated that with words. When I try and fill the array more than half full I want to create a larger structure, copy the current data to that larger structure and then have that larger structure 'replace' the old one that is called from main. Although I have successfully created and copied the data to the new structure; which I can prove by printing the data out from within the function; I am not able to replace the old structure in main. The next book_insert I try inserts to the old, smaller structure not the new, larger one.
I am operating within a constraint whereby I cannot do the resizing / copying / replacing within main; it has to be called from the book_insert function called from main. Additionally I cannot edit void book_insert(dic* s, char* v) (i.e. add double pointers), it has to remain in this format.
#include <stdio.h>
#include <stdlib.h>
struct book {
int size;
int count;
char** words;
};
typedef struct book book;
/* Create empty book, specifying lenght of strings and how many of them */
book* book_init(int wordlen, int maxwords);
/* Add one element into the book */
void book_insert(book* s, char* v);
/* Creates and returns new, bigger book */
book* resize(book* s, book* new);
/* Prints book */
void prints(book* a);
int main(void)
{
book* test;
test = book_init(60, 10);
book_insert(test, "dog");
book_insert(test, "cat");
book_insert(test, "mouse");
book_insert(test, "elephant");
book_insert(test, "snake");
/*The next insert will cause the resize function to trigger*/
book_insert(test, "fish");
/*The resize funtion should cause 'test' to be replaced by a bigger book*/
/*But doesn't as this next print shows*/
printf("But printing from main function means I'm back to %d\n", test->size);
prints(test);
}
void book_insert(book* s, char* v)
{
int i = 0;
while (s->words[i] != NULL ) {
i++;
}
s->words[i] = v;
s->count++;
/*If the book is half full resize is triggered, and should pass back new, bigger book*/
if((s->count * 100 / s->size) > 50) {
book *new_book;
new_book = book_init(60, 20);
s = resize(s, new_book);
printf("Printing from resize function gives me new length of %d\n", s->size);
prints(s);
}
}
book* resize(book* s, book* new)
{
int i;
for (i = 0; i < s->size; i++) {
if (s->words[i] != NULL ) {
new->words[i] = s->words[i];
}
}
return new;
}
book* book_init(int wordlen, int maxwords)
{
int i;
book* new = malloc(sizeof(book));
new->size = maxwords;
new->count = 0;
new->words = (char**) calloc((size_t)new->size, sizeof(char*));
for (i=0; i<new->size; i++) {
new->words[i] = (char*) calloc(wordlen, sizeof(char));
new->words[i] = NULL;
}
return new;
}
void prints(book* a)
{
int i;
for (i = 0; i < a->size; i++) {
printf("Index: %d, word: %s\n", i, a->words[i]);
}
}
I have also attempted this with a pointer swap in a separate function, but this does not seem to work either. In this version I have made book_resize void and instead from dic_insert called the below function, after the resize, with dictionary_swap(&new_book, &s):
void dictionary_swap(book **new, book **old)
{
book *temp = *old;
*old = *new;
*new = temp;
}
This again lets me print out the new larger, structure within the book_insert function, but has no affect on what happens in main.
EDIT ANSWER
This question has been marked as a duplicate, which means I can't answer it myself, however I have since found the answer; I changed the above duplicate swap so that I called dictionary_swap(new_book, s); (no ampersands) on the following code:
void dictionary_swap(book *new, book *old)
{
book temp;
temp = *old;
*old = *new;
*new = temp;
}
In order to modify a pointer inside a function you have to pass the address of the pointer to the function, eg:
void changePtr(char* test) {
test = "Hello";
}
The above will not work because test cannot be returned to the caller, however:
void changePtr(char** test) {
if ( test != NULL ) {
*test = "Hello";
}
}
The above will work because the address of the pointer is passed and it can be de-referenced to change the contents.
Example of calling:
char* ptr;
changePtr(&ptr);
Here is a rewrite of your code implementing the above technique:
#include <stdio.h>
#include <stdlib.h>
typedef struct _book {
int size;
int count;
char** words; //Must allocate space for each pointer before copying to.
} book;
//No need for below, see above:
//typedef struct book book;
/* Create empty book, specifying lenght of strings and how many of them */
book* book_init(int wordlen, int maxwords);
/* Add one element into the book */
void book_insert(book** s, char* v);
/* Creates and returns new, bigger book */
book* resize(book* s, book* new);
/* Prints book */
void prints(book* a);
int main(void) {
book* test = book_init(60, 10);
book_insert(&test, "dog");
book_insert(&test, "cat");
book_insert(&test, "mouse");
book_insert(&test, "elephant");
book_insert(&test, "snake");
/*The next insert will cause the resize function to trigger*/
book_insert(&test, "fish");
/*The resize funtion should cause 'test' to be replaced by a bigger book*/
/*But doesn't as this next print shows*/
printf("But printing from main function means I'm back to %d\n", test->size);
prints(test);
}
void book_insert(book** s, char* v) {
if ( s == NULL || v == NULL ) {
return;
}
(*s)->words = realloc((*s)->words, sizeof(char*) * (++(*s)->count));
(*s)->words[(*s)->count - 1] = v;
/*If the book is half full resize is triggered, and should pass back new, bigger book*/
if((((*s)->count * 100) / s->size) > 50) {
book *new_book;
new_book = book_init(60, 20);
*s = resize(*s, new_book);
}
}
book* resize(book* s, book* new) {
int i;
for (i = 0; i < s->size; i++) {
if (s->words[i] != NULL ) {
new->words[i] = s->words[i];
}
}
printf("Printing from resize function gives me new length of %d\n", new->size);
prints(new);
return new;
}
book* book_init(int wordlen, int maxwords) {
int i;
book* new = calloc(1, sizeof(book));
new->size = maxwords;
return new;
}
void prints(book* a) {
int i;
for (i = 0; i < a->size; i++) {
printf("Index: %d, word: %s\n", i, a->words[i]);
}
}
In the below code, I want to access the structure element that is dynamically allocated inside a function. I have declared structure globally, but allocated inside fun1(). I cannot access the structure element after receiving the structure elements. Kindly help me where I have gone wrong. This works fine if it is in single function. Thanks in advance.
struct s1
{
int a;
char b[10];
} *s2;
int val;
void main()
{
fun1();
fun2();
printf("\n Element a %d",(s1->a));
}
fun1()
{
struct s1 *s2=malloc(sizefof(struct s1)*val);
recv(fd,(void)&s2,sizeof(struct s1),0);
}
fun2()
{
printf("\n Element a %d",(s1->a));
}
There's a lot of confusion here.
Something like this might be what you're after:
struct s1
{
int a;
char b[10];
};
struct s1 * fun1(int val)
{
struct s1 *s = malloc(val * sizeof *s);
if(s != NULL)
recv(fd, s, val * sizeof *s, 0);
return s;
}
void fun2(struct s1 *ptr)
{
printf("First element's a %d", ptr[0].a);
}
int main(void)
{
struct s1 *data = fun1(100);
fun2(data);
printf("First element's a is %d", data[0].a);
return 0;
}
Note that recv() can fail too, and that should be checked. It should, also, perhaps be a read()?
Basically I removed the global variables, and fixed basic errors in allocation and pointer usage.
You have a couple of problems I can see, assuming that int val; was initialized and it's value is reasonable
You don't check for the return value of malloc, it's very unlikely that this is the problem, but that also depends on the value of val.
You pass the address of the pointer to recv.
fun2() makes no sense at all, the code shouldn't even compile.
You redeclare s2 in fun1(), which doesn't seem to be what you want, since you have initially declared s2 global.
The printf() statement in main() is also broken.
Note: in 4 and 5, you should notice that s1 is the name of the struct, which was a very bad choice and pretty much the first source for your confusion.
So fun1() should be re-wirtten
fun1()
{
s2 = malloc(sizefof(*s1) * val);
if (s2 != NULL)
recv(fd, s2, val * sizeof(*s2), 0);
}
and fun2()
fun1()
{
if (s2 != NULL)
printf("\n Element a %d", s2->a);
}
I am working on a Uni assignment here, and I've run into a problem. I am attempting to store a string input at a point inside a struct using a for-loop. Later on I intend to use the pointer to the place where the data was stored to fetch the string. Now the problem is, as I move on inside my for-loop, the address of the point changes as well. This code:
printf("B: %p\n", txt->point);
for(i = 0; i < input_sz; i++)
{
txt->point[i] = input[i];
}
printf("A: %p\n", txt->point);
gives the output:
B: 0x7fc111803200
A: 0x7fc111803265
where B is before-value and A is after-copying value.
Any help debugging this would be very appreciated!
EDIT: Here's some more code:
The struct:
struct text_storage {
char* start;
char* point;
char* end;
} typedef text_t;
Initialization function:
text_t* text_init(void *memory, size_t size)
{
text_t* to_return;
if(size < sizeof(text_t))
{
return NULL;
}
to_return = (text_t*) memory;
to_return->start = to_return;
to_return->end = to_return->start + size;
to_return->point = to_return->start;
printf("Start: %p, point: %p, end: %p, end-start: %d\n", to_return->start, to_return->point, to_return->end, (to_return->end - to_return->start));
return to_return;
}
The text-store method in which the error occurs:
int text_store_entry(text_t *txt, const char *input, size_t input_sz)
{
int to_return;
char* begin = txt->point;
int i;
if(input_sz > (txt->end - txt->point))
{
return -1;
}
printf("Start: %p, point: %p, end: %p, end-start: %d\n", txt->start, txt->point, txt->end, (txt->end - txt->start));
printf("B: %p\n", txt->point);
for(i = 0; i < input_sz; i++)
{
txt->point[i] = input[i];
}
printf("A: %p\n", txt->point);
}
Main-function (testing purposes only):
int main(int argc, char* argv[])
{
void* memory = malloc(10000);
char* a = "hei pa deg din trekkbasun";
text_t* txt;
int memoverwritten;
txt = text_init(memory, 10000);
memoverwritten = text_store_entry(txt, a, (size_t)26);
printf("got through\n");
return 0;
}
The problem most probably is due to the initialization of structures of type struct text_storage. Such structures contain three pointers to text. Each pointer should be initialized, possibly with a malloc. Your text_init function does not do that properly.
In fact, the place where the start pointer is stored overlaps with the first bytes of the memory that you want to use.
I'm guessing that you need a structure like this:
typedef struct text_storage {
char* start;
char* point;
char* end;
char* data;
} text_t;
initialized with a function like this:
text_t text_init(void *memory, size_t size)
{
text_t to_return;
to_return.data = (char *) memory;
to_return.start = to_return.data;
to_return.end = to_return.start + size;
to_return.point = to_return.start;
return to_return;
}
Print txt->point in the loop and see the point at which it changes. I'm guessing it changes when assigning to txt->point[0]. I'm not fully familiar with printf, so I'm not sure what it's printing out for you, but the name of an array references the first location. If printf is printing out a pointer, txt->point[i] is always a char pointer, and printf may be dereferencing txt->point, which will get it the first entry, and then showing the address there, which you do assign when you change the point to input[i].
When should a double indirection be used in C? Can anyone explain with a example?
What I know is that a double indirection is a pointer to a pointer. Why would I need a pointer to a pointer?
If you want to have a list of characters (a word), you can use char *word
If you want a list of words (a sentence), you can use char **sentence
If you want a list of sentences (a monologue), you can use char ***monologue
If you want a list of monologues (a biography), you can use char ****biography
If you want a list of biographies (a bio-library), you can use char *****biolibrary
If you want a list of bio-libraries (a ??lol), you can use char ******lol
... ...
yes, I know these might not be the best data structures
Usage example with a very very very boring lol
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int wordsinsentence(char **x) {
int w = 0;
while (*x) {
w += 1;
x++;
}
return w;
}
int wordsinmono(char ***x) {
int w = 0;
while (*x) {
w += wordsinsentence(*x);
x++;
}
return w;
}
int wordsinbio(char ****x) {
int w = 0;
while (*x) {
w += wordsinmono(*x);
x++;
}
return w;
}
int wordsinlib(char *****x) {
int w = 0;
while (*x) {
w += wordsinbio(*x);
x++;
}
return w;
}
int wordsinlol(char ******x) {
int w = 0;
while (*x) {
w += wordsinlib(*x);
x++;
}
return w;
}
int main(void) {
char *word;
char **sentence;
char ***monologue;
char ****biography;
char *****biolibrary;
char ******lol;
//fill data structure
word = malloc(4 * sizeof *word); // assume it worked
strcpy(word, "foo");
sentence = malloc(4 * sizeof *sentence); // assume it worked
sentence[0] = word;
sentence[1] = word;
sentence[2] = word;
sentence[3] = NULL;
monologue = malloc(4 * sizeof *monologue); // assume it worked
monologue[0] = sentence;
monologue[1] = sentence;
monologue[2] = sentence;
monologue[3] = NULL;
biography = malloc(4 * sizeof *biography); // assume it worked
biography[0] = monologue;
biography[1] = monologue;
biography[2] = monologue;
biography[3] = NULL;
biolibrary = malloc(4 * sizeof *biolibrary); // assume it worked
biolibrary[0] = biography;
biolibrary[1] = biography;
biolibrary[2] = biography;
biolibrary[3] = NULL;
lol = malloc(4 * sizeof *lol); // assume it worked
lol[0] = biolibrary;
lol[1] = biolibrary;
lol[2] = biolibrary;
lol[3] = NULL;
printf("total words in my lol: %d\n", wordsinlol(lol));
free(lol);
free(biolibrary);
free(biography);
free(monologue);
free(sentence);
free(word);
}
Output:
total words in my lol: 243
One reason is you want to change the value of the pointer passed to a function as the function argument, to do this you require pointer to a pointer.
In simple words, Use ** when you want to preserve (OR retain change in) the Memory-Allocation or Assignment even outside of a function call. (So, Pass such function with double pointer arg.)
This may not be a very good example, but will show you the basic use:
#include <stdio.h>
#include <stdlib.h>
void allocate(int **p)
{
*p = (int *)malloc(sizeof(int));
}
int main()
{
int *p = NULL;
allocate(&p);
*p = 42;
printf("%d\n", *p);
free(p);
}
Let’s say you have a pointer. Its value is an address.
but now you want to change that address.
you could. by doing pointer1 = pointer2, you give pointer1 the address of pointer2.
but! if you do that within a function, and you want the result to persist after the function is done, you need do some extra work. you need a new pointer3 just to point to pointer1. pass pointer3 to the function.
here is an example. look at the output below first, to understand.
#include <stdio.h>
int main()
{
int c = 1;
int d = 2;
int e = 3;
int * a = &c;
int * b = &d;
int * f = &e;
int ** pp = &a; // pointer to pointer 'a'
printf("\n a's value: %x \n", a);
printf("\n b's value: %x \n", b);
printf("\n f's value: %x \n", f);
printf("\n can we change a?, lets see \n");
printf("\n a = b \n");
a = b;
printf("\n a's value is now: %x, same as 'b'... it seems we can, but can we do it in a function? lets see... \n", a);
printf("\n cant_change(a, f); \n");
cant_change(a, f);
printf("\n a's value is now: %x, Doh! same as 'b'... that function tricked us. \n", a);
printf("\n NOW! lets see if a pointer to a pointer solution can help us... remember that 'pp' point to 'a' \n");
printf("\n change(pp, f); \n");
change(pp, f);
printf("\n a's value is now: %x, YEAH! same as 'f'... that function ROCKS!!!. \n", a);
return 0;
}
void cant_change(int * x, int * z){
x = z;
printf("\n ----> value of 'a' is: %x inside function, same as 'f', BUT will it be the same outside of this function? lets see\n", x);
}
void change(int ** x, int * z){
*x = z;
printf("\n ----> value of 'a' is: %x inside function, same as 'f', BUT will it be the same outside of this function? lets see\n", *x);
}
Here is the output: (read this first)
a's value: bf94c204
b's value: bf94c208
f's value: bf94c20c
can we change a?, lets see
a = b
a's value is now: bf94c208, same as 'b'... it seems we can, but can we do it in a function? lets see...
cant_change(a, f);
----> value of 'a' is: bf94c20c inside function, same as 'f', BUT will it be the same outside of this function? lets see
a's value is now: bf94c208, Doh! same as 'b'... that function tricked us.
NOW! lets see if a pointer to a pointer solution can help us... remember that 'pp' point to 'a'
change(pp, f);
----> value of 'a' is: bf94c20c inside function, same as 'f', BUT will it be the same outside of this function? lets see
a's value is now: bf94c20c, YEAH! same as 'f'... that function ROCKS!!!.
Adding to Asha's response, if you use single pointer to the example bellow (e.g. alloc1() ) you will lose the reference to the memory allocated inside the function.
#include <stdio.h>
#include <stdlib.h>
void alloc2(int** p) {
*p = (int*)malloc(sizeof(int));
**p = 10;
}
void alloc1(int* p) {
p = (int*)malloc(sizeof(int));
*p = 10;
}
int main(){
int *p = NULL;
alloc1(p);
//printf("%d ",*p);//undefined
alloc2(&p);
printf("%d ",*p);//will print 10
free(p);
return 0;
}
The reason it occurs like this is that in alloc1 the pointer is passed in by value. So, when it is reassigned to the result of the malloc call inside of alloc1, the change does not pertain to code in a different scope.
I saw a very good example today, from this blog post, as I summarize below.
Imagine you have a structure for nodes in a linked list, which probably is
typedef struct node
{
struct node * next;
....
} node;
Now you want to implement a remove_if function, which accepts a removal criterion rm as one of the arguments and traverses the linked list: if an entry satisfies the criterion (something like rm(entry)==true), its node will be removed from the list. In the end, remove_if returns the head (which may be different from the original head) of the linked list.
You may write
for (node * prev = NULL, * curr = head; curr != NULL; )
{
node * const next = curr->next;
if (rm(curr))
{
if (prev) // the node to be removed is not the head
prev->next = next;
else // remove the head
head = next;
free(curr);
}
else
prev = curr;
curr = next;
}
as your for loop. The message is, without double pointers, you have to maintain a prev variable to re-organize the pointers, and handle the two different cases.
But with double pointers, you can actually write
// now head is a double pointer
for (node** curr = head; *curr; )
{
node * entry = *curr;
if (rm(entry))
{
*curr = entry->next;
free(entry);
}
else
curr = &entry->next;
}
You don't need a prev now because you can directly modify what prev->next pointed to.
To make things clearer, let's follow the code a little bit. During the removal:
if entry == *head: it will be *head (==*curr) = *head->next -- head now points to the pointer of the new heading node. You do this by directly changing head's content to a new pointer.
if entry != *head: similarly, *curr is what prev->next pointed to, and now points to entry->next.
No matter in which case, you can re-organize the pointers in a unified way with double pointers.
1. Basic Concept -
When you declare as follows : -
1. char *ch - (called character pointer)
- ch contains the address of a single character.
- (*ch) will dereference to the value of the character..
2. char **ch -
'ch' contains the address of an Array of character pointers. (as in 1)
'*ch' contains the address of a single character. (Note that it's different from 1, due to difference in declaration).
(**ch) will dereference to the exact value of the character..
Adding more pointers expand the dimension of a datatype, from character to string, to array of strings, and so on... You can relate it to a 1d, 2d, 3d matrix..
So, the usage of pointer depends upon how you declare it.
Here is a simple code..
int main()
{
char **p;
p = (char **)malloc(100);
p[0] = (char *)"Apple"; // or write *p, points to location of 'A'
p[1] = (char *)"Banana"; // or write *(p+1), points to location of 'B'
cout << *p << endl; //Prints the first pointer location until it finds '\0'
cout << **p << endl; //Prints the exact character which is being pointed
*p++; //Increments for the next string
cout << *p;
}
2. Another Application of Double Pointers -
(this would also cover pass by reference)
Suppose you want to update a character from a function. If you try the following : -
void func(char ch)
{
ch = 'B';
}
int main()
{
char ptr;
ptr = 'A';
printf("%c", ptr);
func(ptr);
printf("%c\n", ptr);
}
The output will be AA. This doesn't work, as you have "Passed By Value" to the function.
The correct way to do that would be -
void func( char *ptr) //Passed by Reference
{
*ptr = 'B';
}
int main()
{
char *ptr;
ptr = (char *)malloc(sizeof(char) * 1);
*ptr = 'A';
printf("%c\n", *ptr);
func(ptr);
printf("%c\n", *ptr);
}
Now extend this requirement for updating a string instead of character.
For this, you need to receive the parameter in the function as a double pointer.
void func(char **str)
{
strcpy(str, "Second");
}
int main()
{
char **str;
// printf("%d\n", sizeof(char));
*str = (char **)malloc(sizeof(char) * 10); //Can hold 10 character pointers
int i = 0;
for(i=0;i<10;i++)
{
str = (char *)malloc(sizeof(char) * 1); //Each pointer can point to a memory of 1 character.
}
strcpy(str, "First");
printf("%s\n", str);
func(str);
printf("%s\n", str);
}
In this example, method expects a double pointer as a parameter to update the value of a string.
Pointers to pointers also come in handy as "handles" to memory where you want to pass around a "handle" between functions to re-locatable memory. That basically means that the function can change the memory that is being pointed to by the pointer inside the handle variable, and every function or object that is using the handle will properly point to the newly relocated (or allocated) memory. Libraries like to-do this with "opaque" data-types, that is data-types were you don't have to worry about what they're doing with the memory being pointed do, you simply pass around the "handle" between the functions of the library to perform some operations on that memory ... the library functions can be allocating and de-allocating the memory under-the-hood without you having to explicitly worry about the process of memory management or where the handle is pointing.
For instance:
#include <stdlib.h>
typedef unsigned char** handle_type;
//some data_structure that the library functions would work with
typedef struct
{
int data_a;
int data_b;
int data_c;
} LIB_OBJECT;
handle_type lib_create_handle()
{
//initialize the handle with some memory that points to and array of 10 LIB_OBJECTs
handle_type handle = malloc(sizeof(handle_type));
*handle = malloc(sizeof(LIB_OBJECT) * 10);
return handle;
}
void lib_func_a(handle_type handle) { /*does something with array of LIB_OBJECTs*/ }
void lib_func_b(handle_type handle)
{
//does something that takes input LIB_OBJECTs and makes more of them, so has to
//reallocate memory for the new objects that will be created
//first re-allocate the memory somewhere else with more slots, but don't destroy the
//currently allocated slots
*handle = realloc(*handle, sizeof(LIB_OBJECT) * 20);
//...do some operation on the new memory and return
}
void lib_func_c(handle_type handle) { /*does something else to array of LIB_OBJECTs*/ }
void lib_free_handle(handle_type handle)
{
free(*handle);
free(handle);
}
int main()
{
//create a "handle" to some memory that the library functions can use
handle_type my_handle = lib_create_handle();
//do something with that memory
lib_func_a(my_handle);
//do something else with the handle that will make it point somewhere else
//but that's invisible to us from the standpoint of the calling the function and
//working with the handle
lib_func_b(my_handle);
//do something with new memory chunk, but you don't have to think about the fact
//that the memory has moved under the hood ... it's still pointed to by the "handle"
lib_func_c(my_handle);
//deallocate the handle
lib_free_handle(my_handle);
return 0;
}
Hope this helps,
Jason
Strings are a great example of uses of double pointers. The string itself is a pointer, so any time you need to point to a string, you'll need a double pointer.
Simple example that you probably have seen many times before
int main(int argc, char **argv)
In the second parameter you have it: pointer to pointer to char.
Note that the pointer notation (char* c) and the array notation (char c[]) are interchangeable in function arguments. So you could also write char *argv[]. In other words char *argv[] and char **argv are interchangeable.
What the above represents is in fact an array of character sequences (the command line arguments that are given to a program at startup).
See also this answer for more details about the above function signature.
A little late to the party, but hopefully this will help someone.
In C arrays always allocate memory on the stack, thus a function can't return
a (non-static) array due to the fact that memory allocated on the stack
gets freed automatically when the execution reaches the end of the current block.
That's really annoying when you want to deal with two-dimensional arrays
(i.e. matrices) and implement a few functions that can alter and return matrices.
To achieve this, you could use a pointer-to-pointer to implement a matrix with
dynamically allocated memory:
/* Initializes a matrix */
double** init_matrix(int num_rows, int num_cols){
// Allocate memory for num_rows float-pointers
double** A = calloc(num_rows, sizeof(double*));
// return NULL if the memory couldn't allocated
if(A == NULL) return NULL;
// For each double-pointer (row) allocate memory for num_cols floats
for(int i = 0; i < num_rows; i++){
A[i] = calloc(num_cols, sizeof(double));
// return NULL if the memory couldn't allocated
// and free the already allocated memory
if(A[i] == NULL){
for(int j = 0; j < i; j++){
free(A[j]);
}
free(A);
return NULL;
}
}
return A;
}
Here's an illustration:
double** double* double
------------- ---------------------------------------------------------
A ------> | A[0] | ----> | A[0][0] | A[0][1] | A[0][2] | ........ | A[0][cols-1] |
| --------- | ---------------------------------------------------------
| A[1] | ----> | A[1][0] | A[1][1] | A[1][2] | ........ | A[1][cols-1] |
| --------- | ---------------------------------------------------------
| . | .
| . | .
| . | .
| --------- | ---------------------------------------------------------
| A[i] | ----> | A[i][0] | A[i][1] | A[i][2] | ........ | A[i][cols-1] |
| --------- | ---------------------------------------------------------
| . | .
| . | .
| . | .
| --------- | ---------------------------------------------------------
| A[rows-1] | ----> | A[rows-1][0] | A[rows-1][1] | ... | A[rows-1][cols-1] |
------------- ---------------------------------------------------------
The double-pointer-to-double-pointer A points to the first element A[0] of a
memory block whose elements are double-pointers itself. You can imagine these
double-pointers as the rows of the matrix. That's the reason why every
double-pointer allocates memory for num_cols elements of type double.
Furthermore A[i] points to the i-th row, i.e. A[i] points to A[i][0] and
that's just the first double-element of the memory block for the i-th row.
Finally, you can access the element in the i-th row
and j-th column easily with A[i][j].
Here's a complete example that demonstrates the usage:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
/* Initializes a matrix */
double** init_matrix(int num_rows, int num_cols){
// Allocate memory for num_rows double-pointers
double** matrix = calloc(num_rows, sizeof(double*));
// return NULL if the memory couldn't allocated
if(matrix == NULL) return NULL;
// For each double-pointer (row) allocate memory for num_cols
// doubles
for(int i = 0; i < num_rows; i++){
matrix[i] = calloc(num_cols, sizeof(double));
// return NULL if the memory couldn't allocated
// and free the already allocated memory
if(matrix[i] == NULL){
for(int j = 0; j < i; j++){
free(matrix[j]);
}
free(matrix);
return NULL;
}
}
return matrix;
}
/* Fills the matrix with random double-numbers between -1 and 1 */
void randn_fill_matrix(double** matrix, int rows, int cols){
for (int i = 0; i < rows; ++i){
for (int j = 0; j < cols; ++j){
matrix[i][j] = (double) rand()/RAND_MAX*2.0-1.0;
}
}
}
/* Frees the memory allocated by the matrix */
void free_matrix(double** matrix, int rows, int cols){
for(int i = 0; i < rows; i++){
free(matrix[i]);
}
free(matrix);
}
/* Outputs the matrix to the console */
void print_matrix(double** matrix, int rows, int cols){
for(int i = 0; i < rows; i++){
for(int j = 0; j < cols; j++){
printf(" %- f ", matrix[i][j]);
}
printf("\n");
}
}
int main(){
srand(time(NULL));
int m = 3, n = 3;
double** A = init_matrix(m, n);
randn_fill_matrix(A, m, n);
print_matrix(A, m, n);
free_matrix(A, m, n);
return 0;
}
For example, you might want to make sure that when you free the memory of something you set the pointer to null afterwards.
void safeFree(void** memory) {
if (*memory) {
free(*memory);
*memory = NULL;
}
}
When you call this function you'd call it with the address of a pointer
void* myMemory = someCrazyFunctionThatAllocatesMemory();
safeFree(&myMemory);
Now myMemory is set to NULL and any attempt to reuse it will be very obviously wrong.
For instance if you want random access to noncontiguous data.
p -> [p0, p1, p2, ...]
p0 -> data1
p1 -> data2
-- in C
T ** p = (T **) malloc(sizeof(T*) * n);
p[0] = (T*) malloc(sizeof(T));
p[1] = (T*) malloc(sizeof(T));
You store a pointer p that points to an array of pointers. Each pointer points to a piece of data.
If sizeof(T) is big it may not be possible to allocate a contiguous block (ie using malloc) of sizeof(T) * n bytes.
One thing I use them for constantly is when I have an array of objects and I need to perform lookups (binary search) on them by different fields.
I keep the original array...
int num_objects;
OBJECT *original_array = malloc(sizeof(OBJECT)*num_objects);
Then make an array of sorted pointers to the objects.
int compare_object_by_name( const void *v1, const void *v2 ) {
OBJECT *o1 = *(OBJECT **)v1;
OBJECT *o2 = *(OBJECT **)v2;
return (strcmp(o1->name, o2->name);
}
OBJECT **object_ptrs_by_name = malloc(sizeof(OBJECT *)*num_objects);
int i = 0;
for( ; i<num_objects; i++)
object_ptrs_by_name[i] = original_array+i;
qsort(object_ptrs_by_name, num_objects, sizeof(OBJECT *), compare_object_by_name);
You can make as many sorted pointer arrays as you need, then use a binary search on the sorted pointer array to access the object you need by the data you have. The original array of objects can stay unsorted, but each pointer array will be sorted by their specified field.
Why double pointers?
The objective is to change what studentA points to, using a function.
#include <stdio.h>
#include <stdlib.h>
typedef struct Person{
char * name;
} Person;
/**
* we need a ponter to a pointer, example: &studentA
*/
void change(Person ** x, Person * y){
*x = y; // since x is a pointer to a pointer, we access its value: a pointer to a Person struct.
}
void dontChange(Person * x, Person * y){
x = y;
}
int main()
{
Person * studentA = (Person *)malloc(sizeof(Person));
studentA->name = "brian";
Person * studentB = (Person *)malloc(sizeof(Person));
studentB->name = "erich";
/**
* we could have done the job as simple as this!
* but we need more work if we want to use a function to do the job!
*/
// studentA = studentB;
printf("1. studentA = %s (not changed)\n", studentA->name);
dontChange(studentA, studentB);
printf("2. studentA = %s (not changed)\n", studentA->name);
change(&studentA, studentB);
printf("3. studentA = %s (changed!)\n", studentA->name);
return 0;
}
/**
* OUTPUT:
* 1. studentA = brian (not changed)
* 2. studentA = brian (not changed)
* 3. studentA = erich (changed!)
*/
The following is a very simple C++ example that shows that if you want to use a function to set a pointer to point to an object, you need a pointer to a pointer. Otherwise, the pointer will keep reverting to null.
(A C++ answer, but I believe it's the same in C.)
(Also, for reference: Google("pass by value c++") = "By default, arguments in C++ are passed by value. When an argument is passed by value, the argument's value is copied into the function's parameter.")
So we want to set the pointer b equal to the string a.
#include <iostream>
#include <string>
void Function_1(std::string* a, std::string* b) {
b = a;
std::cout << (b == nullptr); // False
}
void Function_2(std::string* a, std::string** b) {
*b = a;
std::cout << (b == nullptr); // False
}
int main() {
std::string a("Hello!");
std::string* b(nullptr);
std::cout << (b == nullptr); // True
Function_1(&a, b);
std::cout << (b == nullptr); // True
Function_2(&a, &b);
std::cout << (b == nullptr); // False
}
// Output: 10100
What happens at the line Function_1(&a, b);?
The "value" of &main::a (an address) is copied into the parameter std::string* Function_1::a. Therefore Function_1::a is a pointer to (i.e. the memory address of) the string main::a.
The "value" of main::b (an address in memory) is copied into the parameter std::string* Function_1::b. Therefore there are now 2 of these addresses in memory, both null pointers. At the line b = a;, the local variable Function_1::b is then changed to equal Function_1::a (= &main::a), but the variable main::b is unchanged. After the call to Function_1, main::b is still a null pointer.
What happens at the line Function_2(&a, &b);?
The treatment of the a variable is the same: within the function, Function_2::a is the address of the string main::a.
But the variable b is now being passed as a pointer to a pointer. The "value" of &main::b (the address of the pointer main::b) is copied into std::string** Function_2::b. Therefore within Function_2, dereferencing this as *Function_2::b will access and modify main::b . So the line *b = a; is actually setting main::b (an address) equal to Function_2::a (= address of main::a) which is what we want.
If you want to use a function to modify a thing, be it an object or an address (pointer), you have to pass in a pointer to that thing. The thing that you actually pass in cannot be modified (in the calling scope) because a local copy is made.
(An exception is if the parameter is a reference, such as std::string& a. But usually these are const. Generally, if you call f(x), if x is an object you should be able to assume that f won't modify x. But if x is a pointer, then you should assume that f might modify the object pointed to by x.)
Compare modifying value of variable versus modifying value of pointer:
#include <stdio.h>
#include <stdlib.h>
void changeA(int (*a))
{
(*a) = 10;
}
void changeP(int *(*P))
{
(*P) = malloc(sizeof((*P)));
}
int main(void)
{
int A = 0;
printf("orig. A = %d\n", A);
changeA(&A);
printf("modi. A = %d\n", A);
/*************************/
int *P = NULL;
printf("orig. P = %p\n", P);
changeP(&P);
printf("modi. P = %p\n", P);
free(P);
return EXIT_SUCCESS;
}
This helped me to avoid returning value of pointer when the pointer was modified by the called function (used in singly linked list).
OLD (bad):
int *func(int *P)
{
...
return P;
}
int main(void)
{
int *pointer;
pointer = func(pointer);
...
}
NEW (better):
void func(int **pointer)
{
...
}
int main(void)
{
int *pointer;
func(&pointer);
...
}
Most of the answers here are more or less related to application programming. Here is an example from embedded systems programming. For example below is an excerpt from the reference manual of NXP's Kinetis KL13 series microcontroller, this code snippet is used to run bootloader, which resides in ROM, from firmware:
"
To get the address of the entry point, the user application reads the word containing the pointer to the bootloader API tree at offset 0x1C of the bootloader's vector table. The vector table is placed at the base of the bootloader's address range, which for the ROM is 0x1C00_0000. Thus, the API tree pointer is at address 0x1C00_001C.
The bootloader API tree is a structure that contains pointers to other structures, which have the function and data addresses for the bootloader. The bootloader entry point is always the first word of the API tree.
"
uint32_t runBootloaderAddress;
void (*runBootloader)(void * arg);
// Read the function address from the ROM API tree.
runBootloaderAddress = **(uint32_t **)(0x1c00001c);
runBootloader = (void (*)(void * arg))runBootloaderAddress;
// Start the bootloader.
runBootloader(NULL);
I have used double pointers today while I was programming something for work, so I can answer why we had to use them (it's the first time I actually had to use double pointers). We had to deal with real time encoding of frames contained in buffers which are members of some structures. In the encoder we had to use a pointer to one of those structures. The problem was that our pointer was being changed to point to other structures from another thread. In order to use the current structure in the encoder, I had to use a double pointer, in order to point to the pointer that was being modified in another thread. It wasn't obvious at first, at least for us, that we had to take this approach. A lot of address were printed in the process :)).
You SHOULD use double pointers when you work on pointers that are changed in other places of your application. You might also find double pointers to be a must when you deal with hardware that returns and address to you.