I have a dynamically allocated array of structures, 'buff'. Each element is a structure that has a few integer variables and a pointer 'buffer_ptr' which points to another dynamically allocated array of structures. The size of both arrays is given as command line input.
int buffer_size;
int user_num;
struct tuple
{
char userID[5];
char topic[16];
int weight;
};
struct buff_ctrl
{
struct tuple* buffer_ptr;
int in;
int out;
int numItems;
int done;
};
The arrays are created and initialized in main() as follows:
int main(int argc, char* argv[])
{
void *status;
pthread_t mapThd;
if(argc != 4)
{
printf("Input format: ./combiner <buffer_size> <number_of_users> <input_file>\n");
return -1;
}
buffer_size = atoi(argv[1]);
user_num = atoi(argv[2]);
struct buff_ctrl *buff = (struct buff_ctrl*)malloc(user_num * sizeof(struct buff_ctrl));
for(int i=0; i<user_num; i++)
{
struct buff_ctrl* curr_buff = (buff + (i*sizeof(struct buff_ctrl)));
struct tuple *ptr = (struct tuple*)malloc(buffer_size * sizeof(struct tuple));
curr_buff->buffer_ptr = ptr;//points to another array
curr_buff->in = 8;
curr_buff->out = 4;
curr_buff->numItems = 7;
curr_buff->done = 0;
printf("%p\n",curr_buff);
}
Then, I need to pass the 'buff' pointer as an argument when creating thread using pthread_create:
pthread_create(&mapThd, NULL, mapper, (void*)buff);
pthread_join(mapThd, &status);
free(buff);
/*end of main*/
My function pointer is as follows:
void* mapper(void *buff)
{
struct buff_ctrl* arr = (struct buff_ctrl *)buff;
struct buff_ctrl* temp_ptr;
printf("######################################################\n");
for(int k=0; k<user_num; k++)
{
/*Printing just to check values */
temp_ptr = arr + (k*sizeof(struct buff_ctrl));
printf("buffer ptr = %p\n", temp_ptr->buffer_ptr);
printf("in = %d\n", temp_ptr->in);
printf("out = %d\n", temp_ptr->out);
printf("numItems = %d\n", temp_ptr->numItems);
}
printf("######################################################\n");
pthread_exit((void*)buff);
}
But, when I print the values of 'buffer_ptr' from the created thread (only one), for ODD number of user_num, there is always ONE element of the array 'buff' which gives garbage value after pthread_create statement! When the values are checked in main itself after removing calls to pthread, it runs fine.
This line
struct buff_ctrl* curr_buff = (buff + (i*sizeof(struct buff_ctrl)));
should be
struct buff_ctrl* curr_buff = buff + i;
buff + i is pointer arithmetic and the compiler already takes the size of the
object pointed to by buff into consideration. By doing i*sizeof(struct buff_ctrl) you are assigning
a pointer that may be after the allocated memory.
As general suggestion:
Don't cast malloc. And instead of using sizeof(<type>), use sizeof *variable, this is more safe, because
it's easier to make mistakes when writing sizeof(<type>).
So:
struct buff_ctrl *buff = malloc(user_num * sizeof *buff);
...
struct tuple *ptr = malloc(buffer_size * sizeof *ptr);
And you don't need to declare a separate pointer, you can do:
for(int i=0; i<user_num; i++)
{
buff[i].buffer_ptr = malloc(buffer_size * sizeof *buff[i].buffer_ptr);
buff[i].in = 8;
buff[i].out = 4;
buff[i].numItems = 7;
buff[i].done = 0;
}
Also you should always check for the return value of malloc. If it returns
NULL, you cannot access that memory.
This is wrong:
struct buff_ctrl* curr_buff = (buff + (i*sizeof(struct buff_ctrl)));
When you do pointer arithmetic, it operates in units of the size of what the pointer points to, so you don't need to multiply by sizeof. As a result, you're effectively multiplying twice and accessing outside the array bounds.
Just treat buff as an array, rather than dealing with pointers.
for(int i=0; i<user_num; i++)
{
struct tuple *ptr = malloc(buffer_size * sizeof(struct tuple));
buff[i].buffer_ptr = ptr;//points to another array
buff[i].in = 8;
buff[i].out = 4;
buff[i].numItems = 7;
buff[i].done = 0;
}
Also, see Do I cast the result of malloc?
You have a fundamental error.
Pointer arithmetics works by adding the offset in multiples of the pointer type, so adding the offset yourself will not work as you apparently expect it to.
If it was a char * pointer then you would need to add the offset manually, increments would be multiplied by one. But in your case increments by n are multiplied by the size of the pointer base type.
There are times when doing pointer arithmetics with the addition notation makes sense, but most of the time it's much clearer to write index notation instead.
I am a newcomer to the world of C. I am self-teaching and would appreciate some help with a couple of questions.
This program is a simplified variation on one written in this example to demonstrate the use of pointers with arrays of structs. The things I'm having trouble getting my head around are:
How the array is incremented in the first for loop. The ++ operator is used directly on the array of structs whereas I would have expected the need to do something more like in_p[i] inside the loop (with i being an int that get incremented).
The way that the comparison is being made in this loop. I didn't think that in_p < &ar[ARSIZE] would be possible since both are of type struct wp_char. What is actually being compared here?
Both the example in the book and my example compile and run.
Thank you.
#include <stdio.h>
#include <stdlib.h>
#define ARSIZE 5
struct wp_char{
char wp_cval;
short wp_font;
short wp_psize;
}ar[ARSIZE];
void infun(struct wp_char *, char cval, int font, int psize);
int main(void)
{
struct wp_char wp_tmp, *lo_indx, *hi_indx, *in_p;
char c[] = {'a','b','c','d','e'};
int i1[] = {2,3,4,5,6};
int i2[] = {7,8,9,10,11};
int i = 0;
for(in_p = ar; in_p < &ar[ARSIZE]; in_p++){
infun(in_p, c[i], i1[i], i2[i]);
i++;
}
int j;
for(j=0;j<ARSIZE;j++)
{
printf("%c\n",c[j]);
printf("%d\n",i1[j]);
printf("%d\n",i2[j]);
puts("\n");
}
exit(0);
}
void infun( struct wp_char *inp, char cval, int font, int psize)
{
`
inp->wp_cval = cval;
inp->wp_font = font;
inp->wp_psize = psize;
return;
}
Question 1:
In C pointer arithmetic, ++ and -- increment and decrement a pointer by the size of the thing being pointed to, not by a byte (or some other aribitrary measure). See e.g. http://www.eskimo.com/~scs/cclass/notes/sx10b.html and http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/BitOp/pointer.html.
Note that this also applies to plain old addition and subtraction too:
float x[10];
float xPtr = &x[0]; // OR could write simply "float xPtr = x;"
xPtr = xPtr + 1; // xPtr now points at x[1]
xPtr = xPtr - 1; // xPtr now points at x[0] again
xPtr = xPtr + 9; // xPtr now points at last item in the array, x[9]
Question 2:
They're not of type struct wp_char, they're of type struct wp_char*, i.e. pointer to that struct. Think of them as a number pointing to a memory location containing one of those structs. You can compare two memory location for equality.
I am trying to have dynamically allocate arrays of structures and perform operations on them but i keep running into segmentation faults. could someone help me out?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void *malloc(size_t size);
typedef struct {
double x;
double y;
} coords;
struct figure {
char fig_name[128];
int coordcount, size_tracker;
coords *pointer;
} fig;
void init_fig(int n, struct figure **point)
{
printf("%u\n", sizeof(coords));
point[n]->pointer = malloc(sizeof(coords) * 20); <-------SEGFAULT
if (point[n]->pointer == NULL){
exit(-1);
}
point[n]->pointer[19].x = 2;
point[n]->pointer[0].x = 1;
point[n]->pointer[0].y = 2;
point[n]->pointer[7].x = 100;
}
int main()
{
int numfigs = 1;
struct figure * point;
point = malloc(sizeof(struct figure) * 16);
point = &fig;
point[1].coordcount = 1;
init_fig(numfigs, &point);
return 0;
}
I labelled where the first seg fault occurs, (used ddd). what i dont get is that i can manipulate point[1] in main but not in any other function.
I agree with #Maxim Skurydin.
Nevertheless I'd like to explain your mistake in some more details.
Reading your init_fig one assumes that the parameter you pass struct figure **point - is actually array of pointers to struct figure. And this function accesses its n'th element.
However in your main you do something else. You allocate an array of struct figure, and your point variable points to its head. Then you take the address of this local variable and call your init_fig.
Here's the problem. init_fig assumes that you pass it an array of pointers, whereas actually this "array" consists of a single element only: the local point variable declared in main.
EDIT:
How to do this properly.
Leave main intact, fix init_fig.
This means that actually there's an array of figure structs. Means - a single memory block, interpreted as an array of consequent structs.
void init_fig(int n, struct figure *point)
{
printf("%u\n", sizeof(coords));
point[n].pointer = malloc(sizeof(coords) * 20); <-------SEGFAULT
if (point[n].pointer == NULL){
exit(-1);
}
point[n].pointer[19].x = 2;
point[n].pointer[0].x = 1;
point[n].pointer[0].y = 2;
point[n].pointer[7].x = 100;
}
Leave init_fig intact. Fix main.
This means that we actually should allocate an array of pointers, every such a pointer should point to an allocated point structure.
int main()
{
int numfigs = 1;
struct figure ** point;
point = malloc(sizeof(struct figure*) * 16);
for (i = 0; i < 16; i++)
point[i] = malloc(sizeof(struct figure));
point[1].coordcount = 1;
init_fig(numfigs, &point);
return 0;
}
You allocate memory and store the pointer in point but then you forget that pointer when you assign &fig to it.
point = malloc(sizeof(struct figure) * 16);
point = &fig;
So, you are essentially trying to write fig[1], that does not make sense.
struct figure * point;
point = malloc(sizeof(struct figure) * 16);
here point is pointer pointing to memory of 16 structures in heap
but in the next line you have done this
point = &fig;
so its memory leak and also point is not pointing to that allocated region anymore
and also init_fig should be like this
void init_fig(int n, struct figure **point)
It's the problem of segfault
Eliminate this line point = &fig;
and modify the function:
void init_fig(int n, struct figure *point)
{
...
point[n].pointer = (coords*) malloc(sizeof(coords) * 20);
...
}
since you should pass an array of structs and not an array of pointers.
Also, add a third parameter to the init_fig function so you can pass the size of the array of points that you want to create. Like :
void init_fig(int n, struct figure *point, int size)
{
...
point[n].pointer = (coords*) malloc(sizeof(coords) * size);
...
}
Therefore, making the function more reusable.
Modify also the call to that function:
init_fig(numfigs, &point); to init_fig(numfigs, point);
I have the following C code which works:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <malloc.h>
int pw = sizeof(char*); // width of pointer (to char)
int num;
int first = 1;
int size = 0;
int incr = 10;
char *(*arr)[]; // pointer to array of pointers to char */
test(char* s, int i)
{
int j;
char *(*newarr)[]; // pointer to array of pointers to char
if (first) { // first time
arr = malloc(pw*incr); // malloc array
first = 0; // skip from now on
size = incr; // save the size
}
if (i >= size) { // out of space
newarr = malloc(pw*(size+incr)); // get incr bigger space
for (j=0; j<size; j++) // copy the elements from the old
(*newarr)[j] = (*arr)[j]; // array to new array
free(arr); // free the old array space
arr = newarr; // point old array to new array
size = size+incr;
};
int len = strlen(s); // length of s
(*arr)[i] = malloc(len+1); // assign pointer to pointer array element
strcpy((*arr)[i], s); // copy s to array
// both arguments must be pointers
printf("%d\t%s\n", i, (*arr)[i]);
};
main()
{
char* s = "this is a string";
for (num=0; num<30; num++) // add 30 pointers to s to *arr
test(s, num);
for (num=0; num<30; num++)
printf("%d\t%s\n", num, (*arr)[num]); // print out what they point to
};
It prints out 'i\tthis is a string' for 'i' from 0 to 29 twice. What I want to do is pass 'arr' from the top of the file as an argument of 'test'. The reason I want to do that is because I want to pass several different arrays all of which are declared the same way. If I make the minimal changes to do that I get:
0 this is a string
Segmentation fault (core dumped)
Here is the output of the diff command which shows the minimal changes:
13c13
< char *(*arr)[]; // pointer to array of pointers to char */
---
> char *(*jarr)[]; // pointer to array of pointers to char */
15c15
< test(char* s, int i)
---
> test(char* s, int i, char *(*arr)[])
52c52
< test(s, num);
---
> test(s, num, jarr);
54,55d53
< for (num=0; num<30; num++)
< printf("%d\t%s\n", num, (*arr)[num]); // print out what they point to
In other words everything is the same except for renaming 'arr' as 'jarr' and passing it to 'test'.
Thanks in advance,
Mike
The trouble occurs when you call:
test(s, num, jarr);
You are passing jarr by value. Inside the function, you are reallocating (the hard way — why not use realloc() which does the copying for you?) the array, but that change does not affect the value of jarr 'in main()' because it was passed by value. The second time through the loop, you are still passing a null pointer to the function, but you are then dereferencing that null pointer, which is bad news.
How to fix?
Fair question...I'm not sure if the old "well, if I want to get to there, I wouldn't start from here" gag passes muster.
The 'simplest' change is to revise the call:
jarr = test(s, num, jarr);
and then 'just' revise the function so that it returns a pointer to an array of character pointers. That is a very esoteric function. My brain's not awake (insufficient caffeine), so I used an intermediate typedef to get around the problem of how to write the function declaration and definition:
typedef char *(ArrayString[]);
ArrayString *test3(char *s, int i, char *(*arr)[]);
ArrayString *test3(char *s, int i, char *(*arr)[]) { (*arr)[i] = s; return arr; }
It compiles without warnings; that isn't a guarantee that it's correct.
The primary alternative is to pass a pointer to a pointer to an array of char pointers to the function, which is even more esoteric.
However, both of these are 'starting from here' solutions. You'd do better, on the whole, to devise a different way of handling things. Pointers to arrays are certainly a part of C, but they are at the outer edges of C and you should generally assume that if your design calls for their use, then your design is probably not the best. You should use a simpler char ** (or, perish the thought, char ***; triple indirection is best avoided too, but that isn't always possible).
You seem to have misunderstood how arrays and pointers works. Lets say you want a dynamic array of strings, that is basically a pointer to a pointer of char:
char **arr = NULL;
To allocate memory for that you do e.g.
arr = malloc(sizeof(char *) * current_size);
Now you have an "array" of character pointers. Lets say you want each of these to be a specific string str:
for (int i = 0; i < current_size; i++)
{
arr[i] = strdup(str);
}
Oh, now you need to increase the number of strings, all initialized to the same string as before:
size_t new_size = current_size + 10;
arr = realloc(arr, sizeof(char *) * new_size);
for (int i = current_size; i < new_size)
{
arr[i] = strdup(str);
}
The problem now is that you want to do all of the above in a separate function. It's first now that you have to add another indirection.
I think you can do a double check on the first malloc value assigned to jarr both in the test(s, 0, jarr) and out of the test(s, 0, jarr); the jarr assignement is not successful since you change the pointer value in the passing by value.
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.