I'm newbie in C language. If I assign the address of local variable to global pointer, What happens? Like,
#include <stdio.h>
void func();
int *ptr;
int main()
{
func();
}
void func()
{
int i = 0;
ptr = &i;
}
Is it correct way to assign the address of local variable to global pointer?
It just does what you do, there is nothing wrong about it, except that it probably is not what you want.
So it just assigns the address of i to ptr at the point you assign it. When you leave func this pointer gets invalid.
Note This behaviour is fully defined: The address of i at the place you assign it to the global variable is defined, and so you can assign it. The problem only comes into play later, when you try to dereference the variable after you left the function func. As long as you only use the global variable in func there is no problem in this (except that a global variable is really meaningless).
At this point, this variable doesn't exist anymore. And it is very likely that you either get a segfault or at least you get some strange numbers (because you have overwritten the old stack frame with some other values) in this case.
Just as a side note: What I mean with this stack frame thing. You can try this code on most compilers (without optimizations!)
#include <stdio.h>
int *ptr;
void f1() {
int i = 0;
ptr = &i;
}
void f2() {
int i = 1;
}
int main() {
f1();
printf("%d\n", *ptr);
f2();
printf("%d\n", *ptr);
}
Without optimizations, this will most probably print
0
1
Because the variable i will have the same address when calling f1 and f2 from main().
With optimizations, the call to f2() will be optimized.
Still: This is undefined behavior and must not be done.
Your syntax is correct, but the local variable ceases to exist because it belongs within the scope of the function call's code block. To resolve this, one option is to make the local variable static:
#include <stdio.h>
void func();
int *ptr;
int main()
{
func();
}
void func()
{
static int i = 0;
ptr = &i;
}
Another option would be to allocate new memory within the function call, setting the global pointer to the address of that newly allocated memory:
#include <stdio.h>
void func();
int *ptr = NULL;
int main()
{
func();
}
void func()
{
if(ptr != NULL)
free(ptr);
int *i = (int *)malloc(sizeof(int));
ptr = i;
}
What you've got is syntactically correct, and the code as written is semantically valid (but since ptr is never used, it is a bit pointless).
If you access ptr when it contains a pointer that has gone out of scope, you get undefined behaviour.
However, consider a slightly larger code fragment. Here, the code that sets ptr calls a function that uses ptr, and the variable that ptr points to is still defined, so there is no problem using the pointer.
#include <stdio.h>
void func(void);
void use_pointer(void);
int *ptr;
int main(void)
{
func(); // NB: argument not allowed with prototype!
int i = 20;
printf("%s: A %d\n", __func__, i);
ptr = &i;
use_pointer();
printf("%s: B %d\n", __func__, i);
}
void func(void)
{
int i = 0;
printf("%s: A %d\n", __func__, i);
ptr = &i;
use_pointer();
printf("%s: B %d\n", __func__, i);
}
void use_pointer(void)
{
printf("ptr = %p; *ptr = %d\n", (void *)ptr, *ptr);
*ptr = 42;
}
This is legitimate code — though using global variables is something you should generally avoid and perfectly well could avoid.
Sample output:
func: A 0
ptr = 0x7fff55be74ac; *ptr = 0
func: B 42
main: A 20
ptr = 0x7fff55be74cc; *ptr = 20
main: B 42
#include <stdio.h>
struct Analysis {
int lnlen;
int arr[2];
char* name;
};
int main()
{
struct Analysis ana_space[2];
char *ptr = (void*) &ana_space;
ana_space[0].lnlen = 0;
ana_space[0].arr[0] = 1;
ana_space[0].arr[1] = 2;
ana_space[0].name = "Peter";
printf("\n%d\n", *ptr); // print 0;
*ptr = 10; // how to use memcpy here;
printf("\n%d\n", *ptr); // print 10;
ptr = ptr + sizeof(int); // advance pointer by int;
printf("\n%d\n", *ptr); // print 1;
ptr = ptr + 2*sizeof(int); // advance pointer by 2 ints;
printf("\n%s\n", *ptr); // print "Peter"; --------------not work
//*ptr = "Jim"; // how to assign new name "Jim" into that memory;
return 0;
}
Output:
0
10
1
(null)
I want to use char * as pointer to go through memory address to get some data and also store value into memory.
For int and int array, it works fine, but not for the string.
How to print the string and store new string value into memory?
Your approach is not portable. It will be better to use offsetof to make sure that you can reliably point to the addresses of the members of a struct.
int main()
{
struct Analysis ana_space[2];
char *ptr = (void*) &ana_space;
size_t offset1 = offsetof(struct Analysis, arr);
size_t offset2 = offsetof(struct Analysis, name);
ana_space[0].lnlen = 0;
ana_space[0].arr[0] = 1;
ana_space[0].arr[1] = 2;
ana_space[0].name = "Peter";
// advance pointer to point to arr.
ptr = ptr + offset1;
// advance pointer to point to name
ptr = ptr + (offset2-offset1);
// Cast the pointer appropriately before dereferencing.
printf("\n%s\n", *(char**)ptr);
// how to assign new name "Jim" into that memory;
*(char**)ptr = "Jim";
printf("\n%s\n", *(char**)ptr);
return 0;
}
Your use of:
printf("\n%d\n", *ptr); // print 0;
*ptr = 10; // how to use memcpy here;
printf("\n%d\n", *ptr); // print 10;
and the expected output is flawed. It works only with little endian systems. I suggest using:
printf("\n%d\n", *(int*)ptr);
*(int*)ptr = 10;
printf("\n%d\n", *(int*)ptr);
The code you presented could cause undefined behavior due to padding and representations of types, which are implementation-defined.
After you increment the pointer ptr here:
ptr = ptr + 2*sizeof(int);
the pointer ptr points to the member name of the struct Analysis. If you dereference the pointer ptr, you get the type char and thus a single byte. This byte does not represent a pointer to the string.
The pointer ptr will have to be cast to the type pointer to a pointer to char, and then dereferenced so the correct and full value of the member name will be obtained.
That resulting value is a pointer to the string "Peter".
ANSI C has a macro called offsetof() found in stddef.h that gives a more sure way of calculating the pointer offset of a member within a struct. Here, we can get the the address of the name member in ana_space[0] directly.
ptr = (char*) &ana_space + offsetof(struct Analysis, name);
This takes out any guess work on padding.
This pointer then has to be properly cast to print the contents of name:
printf("%s\n", *(char**) ptr);
I have a program which creates an array or struct and go through it for processing. Initially it initialize the array with the defined nyumber of elements. Then for some number of element in array, the name is assigned.
I pretend the code that is equal to my scenario which is tested in codebloc and get the similar error. The problem is described in comments.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct _car {
char *name;
int year;
} Car;
char *getCarName(char *name, int var);
void processCar();
void printCars(Car car[]);
int INCREMENT = 10;
int main(void)
{
processCar();
return 0;
}
void processCar()
{
// create car array with amount of INCREMENT
Car CAR_ARRAY[INCREMENT];
int a=0;
// This function assign name for Car amount 10 less than INCREMENT
while (a<INCREMENT - 2) {
char *carName;
carName = getCarName(&carName, a);
CAR_ARRAY[a].name = malloc(strlen(carName) + 1);
strcpy(CAR_ARRAY[a].name, carName);
a++;
}
printCars(CAR_ARRAY);
}
void printCars(Car car[])
{
printf("IN Car \n");
int a = 0;
// when try to call name for car amount equals to INCREMENT program terminates.
while(a<INCREMENT) {
if (car[a].name != NULL) // checking if NULL
printf("Car Name : %d -> %s\n", a, car[a].name);
a++;
}
}
char *getCarName(char *name, int var)
{
name = "Toyota";
return name;
}
What is the right way to check the struct value on struct array whether it can be called?
EDIT
I created a hack to do this as follows.
// added these right after creating array
for (a = 0; a < INCREMENT; a++)
CAR_ARRAY[a].name = NULL;
I dont know if it is a standard way or not. pls advice.
You are checking for NULL before printing, which is a good idea, but it doesn't help you here, because your last two cars are uninitialised and likely contain garbage: The name pointer is not NULL, but doesn't point to a valid address either. Segmentation violation ensues.
You should initialise all cars, not only INCREMENT - 2. Alternatively, you could initialise your cars to zero by calling memset before your initialisation:
memset(CAR_ARRAY, 0, sizeof(Car) * INCREMENT);
As an aside, the way you deal with getCarName is rather shaky as well. At the moment, your name is a pointer to a string literal. Your local variable carName does a half-hearted double duty: You try to pass it by reference (but essentially you don't) and you also return it.
Basically, you could do this in one of two ways. The easier one here is to return a pointer. in that case, you don't have to pass any string:
char *getCarName(int var)
{
static char *names[3] = {"Toyota", "Dodge", "Peugeot"};
return names[var % 3];
}
and call it like so:
char *carName = getCarName(&carName, a);
Alternatively, you could pass a char pointer by reference, i.e. as pointer to pointer to char. In that case, you don't have to return anything:
void getCarName(char **name, int var)
{
static char* names[3] = {"Toyota", "Dodge", "Peugeot"};
*name = names[var % 3];
}
Call it like so:
char *carName;
getCarName(&carName, a);
There are other scenarios here, for example if you just pass a char pointer and have getCarName fill it, but I'll leave that for now - it would make everything even more complicated.
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.
This question already has answers here:
C Programming: malloc() inside another function
(9 answers)
Closed 5 years ago.
I would like to know the technical reason(in terms of memory) why this piece of code will not work:
#include <stdio.h>
#include <stdlib.h>
int* fun(int*);
int main()
{
int a=5;
int* ptr;
// ptr=(int*)malloc(sizeof(int));
fun(ptr);
a=*ptr;
printf("\n the val of a is:%d",a);
return 0;
}
void fun(int* ptr)
{
ptr = (int*)malloc(sizeof(int));
*ptr = 115;
}
Why will this not work? I thought that the heap(more importantly the addresses) is common to all the function's variables in the stack .
Also, why would this work.
If i comment the memory allocation inside the function fun and uncomment the one in main . It works fine.
In C, everything is passed by value.
What you are passing to fun() is a copy of the pointer you have in main().
That means the copy of ptr is aimed at the allocated memory, and that memory set to 115.
The ptr in main() still points at an undefined location because it has never been assigned.
Try passing a pointer to the pointer, so that within fun() you have access to the pointer itself:
#include <stdio.h>
#include <stdlib.h>
int* fun(int**); // <<-- CHANGE
int main()
{
int a=5;
int* ptr;
// ptr=(int*)malloc(sizeof(int));
fun(&ptr); // <<-- CHANGE
a=*ptr;
printf("\n the val of a is:%d",a);
return 0;
}
int* fun(int** another_ptr) // <<-- CHANGE
{
*another_ptr = (int*)malloc(sizeof(int)); // <<-- CHANGE
**another_ptr = 115; // <<-- CHANGE
return *another_ptr;
}
The other option would be to make fun() actually return the updated pointer (as advertised), and assign this to ptr:
#include <stdio.h>
#include <stdlib.h>
int* fun(int*);
int main()
{
int a=5;
int* ptr;
// ptr=(int*)malloc(sizeof(int));
ptr = fun(ptr); // <<-- CHANGE
a=*ptr;
printf("\n the val of a is:%d",a);
return 0;
}
int* fun(int* another_ptr)
{
another_ptr = (int*)malloc(sizeof(int));
*another_ptr = 115;
return another_ptr; // <<-- CHANGE
}
Edit: I renamed the variable in fun() to make it clear that it is different from the one you use in main(). Same name doesn't mean anything here.
The fun() function parameter is a copy of the variable you passed into fun(). So when you do:
ptr = (int*)malloc(sizeof(int));
*ptr = 115;
you only change that copy. You should change the function signature:
int* fun(int** ptr)
{
*ptr = (int*)malloc(sizeof(int));
**ptr = 115;
}
and change how you call it accordingly.
You are confused about several things here, but one easy way of writing the function is:
int * fun()
{
int * ptr = (int*)malloc(sizeof(int));
* ptr = 115;
return ptr;
}
You are now responsible for freeing the memory, so in main():
int * ip = fun();
printf( "%d", * ip );
free( ip );
The alternative is to pass the address of apointer (a pointer to a pointer) to the function:
void fun( int ** pp )
{
* pp = (int*)malloc(sizeof(int));
** pp = 115;
}
then your code in main() looks like:
int * ip;
fun( & ip );
printf( "%d", * ip );
free( ip );
I think you can see that the first function is simpler to use.
You need to pass the address of the pointer in main if you want to change it:
fun(&ptr);
(and change fun appropriately, of course)
At the moment, it's changing the local variable ptr inside the function, and of course that change doesn't magically appear anywhere else.
You're passing the ptr by value to fun. fun will recieve a copy of ptr which will be modified. You need to pass ptr as int**.
void fun(int** ptr)
{
*ptr = (int*)malloc(sizeof(int));
**ptr = 115;
}
and call it with:
fun(&ptr);
(I also removed the return value from fun since it wasn't used)
The variable int* ptr is passed by value to the function fun. So the value assigned to ptr inside the function using ptr = (int*)malloc(sizeof(int)); will not be reflected outside the function. So when you do a = *ptr; in main() you are trying to use an un-initialized pointer. If you want to to reflect the changes done to ptr outside the function then you need to change the signature of fun to fun(int** ptr) and do *ptr = (int*)malloc(sizeof(int));
Remember that if you want a function to modify the value of an argument, you must pass a pointer to that argument. This applies to pointer values; if you want a function to modify a pointer value (not what the pointer points to), you must pass a pointer to that pointer:
void fun (int **ptr)
{
/**
* Do not cast the result of malloc() unless you are
* working with a *very* old compiler (pre-C89).
* Doing so will supress a valuable warning if you
* forget to include stdlib.h or otherwise don't have
* a prototype for malloc in scope.
*
* Also, use the sizeof operator on the item you're
* allocating, rather than a type expression; if you
* change the base type of ptr (say from int to long),
* then you don't have to change all the corresponding
* malloc() calls as well.
*
* type of ptr = int **
* type of *ptr = int *
* type of **ptr = int
*/
*ptr = malloc(sizeof **ptr);
*ptr = 115;
}
int main(void)
{
int *p;
fun(&p);
printf("Integer value stored at %p is %d\n", (void *) p, *p);
return 0;
}
BTW, you have a type mismatch in your example; your initial declaration of fun returns an int *, but the definition returns void.