I am reading the source code of berkeley-abc. There is one code snippet, which is the following.
char *Aig_MmFixedEntryFetch(Aig_MmFixed_t *p)
{
char *pTemp;
int i;
// check if there are still free entries
if (p->nEntriesUsed == p->nEntriesAlloc)
{ // need to allocate more entries
assert(p->pEntriesFree == NULL);
if (p->nChunks == p->nChunksAlloc)
{
p->nChunksAlloc *= 2;
p->pChunks = ABC_REALLOC(char *, p->pChunks, p->nChunksAlloc);
}
p->pEntriesFree = ABC_ALLOC(char, p->nEntrySize * p->nChunkSize);
p->nMemoryAlloc += p->nEntrySize * p->nChunkSize;
// transform these entries into a linked list
pTemp = p->pEntriesFree;
for (i = 1; i < p->nChunkSize; i++)
{
*((char **)pTemp) = pTemp + p->nEntrySize;
pTemp += p->nEntrySize;
}
// set the last link
*((char **)pTemp) = NULL;
// add the chunk to the chunk storage
p->pChunks[p->nChunks++] = p->pEntriesFree;
// add to the number of entries allocated
p->nEntriesAlloc += p->nChunkSize;
}
// incrememt the counter of used entries
p->nEntriesUsed++;
if (p->nEntriesMax < p->nEntriesUsed)
p->nEntriesMax = p->nEntriesUsed;
// return the first entry in the free entry list
pTemp = p->pEntriesFree;
p->pEntriesFree = *((char **)pTemp);
return pTemp;
}
I can not understand what happened when executing "*((char **)pTemp) = pTemp + p->nEntrySize".
So I write a example like that, which is the following.
#include <stdio.h>
int main() {
char* pTemp = "aaaaaaaaaa";
printf("%s 1\n", pTemp);
*((char**)pTemp) = pTemp+2;
printf("%s 2\n", pTemp);
return 0;
}
The example can be compiled to one executable file, there is no error. However, when executing this executable file, there is segmentation fault, the executing result is as following.
aaaaaaaaaa 1
zsh: segmentation fault (core dumped)
And why not *pTemp = pTemp + ... ?
I don't want to critizice the code, but that cast (char **) and the naked address approach seem a bit...creative.
With a slightly different concept this can be done cleaner. Instead of a pointer to byte (char), I have pointer to address (to void). One indirection more, but no fake basic type (void * instead of char (no star)):
#include <stdio.h>
#include <stdlib.h>
int main() {
void **vpa = malloc(4096); // arr. of ptrs (to void)
void **currp = vpa;
int skip_elem = 5; // 5 * sizeof *vpa -> 5*8 bytes entry size
for (int i = 0; i < 7; i++) {
*currp = currp + skip_elem;
currp += skip_elem;
}
for (int i = 0; i < 50; i++)
printf("%p %p\n", vpa+i, vpa[i]);
free(vpa);
}
Output is seven blocks filled like this:
0x55f9dc7922a0 0x55f9dc7922c8
0x55f9dc7922a8 (nil)
0x55f9dc7922b0 (nil)
0x55f9dc7922b8 (nil)
0x55f9dc7922c0 (nil)
0x55f9dc7922c8 0x55f9dc7922f0
0x55f9dc7922d0 (nil)
0x55f9dc7922d8 (nil)
0x55f9dc7922e0 (nil)
0x55f9dc7922e8 (nil)
0x55f9dc7922f0 0x55f9dc792318
0x55f9dc7922f8 (nil)
0x55f9dc792300 (nil)
0x55f9dc792308 (nil)
0x55f9dc792310 (nil)
0x55f9dc792318 0x55f9dc792340
...
With char ** I get compiler warning about char * having a char ** assigned (in *currp = ).
*((char **)pTemp) = pTemp + p->nEntrySize; says to put the address that is p->nEntrySize bytes beyond pTemp into a char * located at pTemp.
pTemp points to some memory that is being organized into some number of “entries.” The code is making it into a linked list by putting a pointer (a char *) at the beginning of each entry that points to the next entry.
Given the pointer pTemp that points to the current entry and an entry size of p->nEntrySize bytes, pTemp + p->nEntrySize is the address of the next entry. Then (char **)pTemp converts the char * pTemp to a char **, a pointer to a char *. So the assignment says “Put the address computed for pTemp + p->nEntrySize at the address pointed to by pTemp.
This is different from pTemp += p->nEntrySize;, which says to change pTemp, rather than to change the memory pTemp points to.
Assuming there is a function like this
int foo (char** str, int x)
{
char* p = *str + x;
foo2(&p); // declared as int foo2 (char** );
}
(oversimplified of course, the real function is recursive and much more complicated)
I've tried to do this:
int foo (char** str, int x)
{
foo2(&(*str + x));
}
But the compiler failed with error:
error: lvalue required as unary '&' operand
Why did the compiler shoot out with this error and how do I pass the pointer to a pointer to string x-byte(s) forwards, without declaring a variable and use its own address?
EDIT
Seems like there is some misunderstanding so I will post a complete simulation of what I want to achieve.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* string = "This is a sample string.";
char* ptr;
int randomizer;
int receive_string (char* buffer, int size) // recv
{
int i = 0;
if(ptr == NULL)
ptr = string;
for(i = 0; *ptr != '\0' && i < size; ptr++)
{
if(randomizer == 2)
{
randomizer++;
break;
}
buffer[i] = *ptr;
i++;
randomizer++;
}
if(*ptr == '\0')
{
buffer[i] = *ptr;
i++;
}
return i;
}
int read_string (char* *buffer, int size, int alloc)
{
int bytes = 0;
printf("Reading string..\n");
if(*buffer == NULL && alloc == 1)
{
printf("Allocating buffer..\n");
*buffer = calloc(size, sizeof(char));
}
bytes = receive_string(*buffer, size);
if(bytes == (-1))
{
return(-1);
}
if(bytes == 0)
{
return 0;
}
if(bytes < size)
{
char* p = *buffer + bytes;
//int temp = read_string(&p, size - bytes, 0); // works
//int temp = read_string(&(char *){&(*buffer)[bytes]}, size - bytes, 0); // works
int temp = read_string(buffer + bytes, size - bytes, 0); // doesn't work
if(temp > 0)
bytes += temp;
else return bytes;
}
return bytes;
}
int main()
{
char* buffer = NULL;
int bytes = read_string(&buffer, strlen(string) + 1, 1);
printf("[%u][%s]\n", bytes, buffer);
if(buffer)
free(buffer);
return 0;
}
The randomizer is the dumbest quickie to "simulate" a recv() that can not receive all bytes. This implementation simulates recv() but instead of reading from a socket queue it reads from a global string.
(*str + x) is not an lvalue as it is a temporay value that does not have an address so you cannot take its address with &. Even if the compiler stored the value in a temporary variable in RAM so its address could be taken how would you reference its value afterwards if foo2() modified the contents of the temporay variable.
Therefore you need to store the value in a temporary variable yourself.
if you want to pass the pointer to pointer to the particular char
foo2(&(char *){&(*str)[x]});
or
I think the following code is what you are trying to do. For kicks, I made it recursive and tested it with the alphabet for a string. Variables cnt and lmt need to be global. It will show a shrinking string if you run it. Just be sure to keep p and lmt small enough to not overflow the string.
void foo(char *s, int p) {
cnt++;
printf("%s\n", s);
if(cnt != lmt) foo(&s[p], p);
}
There is my C code, it is a leetcode problem, and I got "Runtime Error". So I recompile in VS2013, the problem is free(++tmp), why? I can't get it, I writen C code like that, just want to known more things about pointer.
#include <stdio.h>
#include <stdlib.h>
/* Add binary.
* a = "11", b = "1"
* result = "100"
*/
char *add_binary(char *a, char *b);
int main()
{
printf("%s\n", add_binary("10", "1"));
printf("%s\n", add_binary("1111", "1111"));
return 0;
}
char *add_binary(char *a, char *b)
{
int alen = 0, blen = 0, sum = 0;
int len;
char *tmp, *result;
while(*a++) alen++;
while(*b++) blen++;
a -= 2;
b -= 2;
len = alen > blen ? alen : blen;
tmp = (char *)malloc(len*sizeof(char));
printf("%p\n", tmp);
while(*a || *b){
if(*a){
sum += *a - '0' + 0;
a--;
}
if(*b){
sum += *b - '0' + 0;
b--;
}
if(sum > 1){
*tmp++ = 3 == sum ? '1' : '0';
sum = 1;
} else {
*tmp++ = 1 == sum ? '1' : '0';
sum = 0;
}
}
*tmp = '\0';
len += 1 == sum ? 1 : 0;
result = (char *)malloc(len*sizeof(char));
if(1 == sum){
*result++ = '1';
}
while(*(--tmp)){
*result++ = *tmp;
}
*result = '\0';
printf("%p\n", tmp);
free(++tmp);
tmp = NULL;
return (result-len);
}
You can only pass to free the resulting pointer value of malloc:
tmp = (char *)malloc(len*sizeof(char));
then
free(tmp);
is OK.
But free(++tmp) or free(tmp + 42) is not OK and invokes undefined behavior.
Stop modifying the mallocated pointer before freeing it. If you want to use pointer arithmetic, eg '*tmp++', then keep a copy of the original so that the space can be freed.
I have no clue why you would do 'free(++tmp);'. It makes no sense though, by that time, you've already totally shagged up tmp by incrementing it in the while loop:(
Edit: BTW, you've screwed 'result' as well. You are returning a malloced and bodged pointer that cannot be correctly freed by the caller.
Whatever 'clever' thing you are attempting with the pointer manipulations, stop it. It's too easy to get it wrong!
Here is a slightly more detailed response from the other answer.
The pointer is an address for which you allocate memory. When you pass in an address to free that has been previously malloc'd there is no problem. The problem is that you are not passing in the address of a malloc'd space. You are passing in the address of something that is potentially within a malloc'd space and as such cannot be freed.
free expects its argument to be the same pointer value that was returned from a previous malloc or realloc call. If you modify that pointer value before passing it to free, then the behavior is undefined and Bad Things can happen (this is why it appears to work for one platform and breaks on another; in truth, it's broken for both).
You'll need to preserve the pointer values returned from your malloc calls:
char *tmp, *tmpOrig;
...
tmp = tmpOrig = malloc(len * sizeof *tmpOrig); // note no cast, operand of sizeof
...
/**
* modify tmp to your heart's desire
*/
...
free( tmpOrig );
You'll need to do the same thing for result.
You should not cast the result of malloc, unless you are working with a pre-C89 compiler. It's unnecessary, and under C89/C90 compilers can mask a bug.
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.