I need to write a function that creates a double pointer using malloc.
This is how I declared my double pointer as I normally would:
double **G; //Create double pointer to hold 2d matrix
*G = malloc(numNodes * sizeof(double*));
for(i = 0; i < numNodes; i++)
{
G[i] = malloc(numNodes*sizeof(double));
for (j = 0; j < numNodes; j++)
{
G[i][j] = 0;
}
}
Now I tried replacing it with:
double **G;
mallocDoubleArr(G, numNodes);
With the function being:
void mallocDoubleArr(double **arr, int size)
{
int i, j;
*arr = malloc(size * sizeof(double*));
for(i = 0; i < size; i++)
{
arr[i]= malloc(size*sizeof(double));
for (j = 0; j < size; j++)
{
arr[i][j] = 0;
}
}
}
Why doesn't this work?
You need one more "indirection", in other words pass G by reference like a pointer to a pointer to a pointer to float:
void mallocDoubleArr(double ***arr, int size);
And then call it as
mallocDoubleArr(&G, numNodes);
Modify mallocDoubleArr accordingly, like for example
(*arr)[i] = malloc(size*sizeof(double));
For starters, you need to change the line
*G = malloc(numNodes * sizeof(double*));
to
G = malloc(numNodes * sizeof(double*));
(you can't dereference a pointer safely until you've assigned something to it.)
Secondly, your function modifies the pointer passed in, so you need a pointer to it. Your signature should instead by
void mallocDoubleArr(double ***arr, int size)
and you will need to add the relevant indirections in your code to access the pointer that the pointer is pointing to.
A lot of confusion for beginners working with pointers comes from, in my opinion, thinking that they are something different than regular old variables. Pointers, like ints, floats, etc. are just variables that live on the stack: they have addresses, and they are passed to functions the same way. If you want to change a variable (int, float, pointer, etc) in a function, you need to pass a pointer to it. There is no difference in this regard.
C is call-by-value. In
double **G;
mallocDoubleArr(G, numNodes);
you are passing an uninitialized variable to mallocDoubleArr. It may be zero, it may be something else, but it almost certainly isn't something that mallocDoubleArr can assign to.
We can change the code, and the function's definition, to pass in G's address, but then you're dealing with yet another level of pointer. That might make it harder to understand the code. If that really isn't a requirement, I'd propose instead to have mallocDoubleArr return a double**.
double **G;
G = mallocDoubleArr(numNodes);
double **mallocDoubleArr(int size)
{
int i, j;
double **arr;
arr = (double **) malloc(size * sizeof(double *));
/* continue with old code */
return arr;
}
[edit: bstamour's post was made while I was writing mine. Sorry for any overlap.]
I use for matrix operations code like following for allocating and freeing.
int **inputMatrix, i, j;
Grid myGrid = *grid;
inputMatrix = (int *) calloc (myGrid.num_nodes, sizeof(int*));
for(i=0; i < myGrid.num_nodes; i++){
inputMatrix[i] = calloc(myGrid.num_nodes, sizeof(int));
for(j=0;j<myGrid.num_nodes;j++){
inputMatrix[i][j] = 0;
}
};
for(i=0; i < myGrid.num_nodes; i++){
free(inputMatrix[i]);
}
free (inputMatrix);
Related
I dynamically allocated memory for 3D array of pointers. My question is how many pointers do I have? I mean, do I have X·Y number of pointers pointing to an array of double or X·Y·Z pointers pointing to a double element or is there another variant?
double*** arr;
arr = (double***)calloc(X, sizeof(double));
for (int i = 0; i < X; ++i) {
*(arr + i) = (double**)calloc(Y, sizeof(double));
for (int k = 0; k < Y; ++k) {
*(*(arr+i) + k) = (double*)calloc(Z, sizeof(double));
}
}
The code you apparently intended to write would start:
double ***arr = calloc(X, sizeof *arr);
Notes:
Here we define one pointer, arr, and set it to point to memory provided by calloc.
Using sizeof (double) with this is wrong; arr is going to point to things of type double **, so we want the size of that. The sizeof operator accepts either types in parentheses or objects. So we can write sizeof *arr to mean “the size of a thing that arr will point to”. This always gets the right size for whatever arr points to; we never have to figure out the type.
There is no need to use calloc if we are going to assign values to all of the elements. We can use just double ***arr = malloc(X * sizeof *arr);.
In C, there is no need to cast the return value of calloc or malloc. Its type is void *, and the compiler will automatically convert that to whatever pointer type we assign it to. If the compiler complains, you are probably using a C++ compiler, not a C compiler, and the rules are different.
You should check the return value from calloc or malloc in case not enough memory was available. For brevity, I omit showing the code for that.
Then the code would continue:
for (ptrdiff_t i = 0; i < X; ++i)
{
arr[i] = calloc(Y, sizeof *arr[i]);
…
}
Notes:
Here we assign values to the X pointers that arr points to.
ptrdiff_t is defined in stddef.h. You should generally use it for array indices, unless there is a reason to use another type.
arr[i] is equivalent to *(arr + i) but is generally easier for humans to read and think about.
As before sizeof *arr[i] automatically gives us the right size for the pointer we are setting, arr[i].
Finally, the … in there is:
for (ptrdiff_t k = 0; k < Y; ++k)
arr[i][k] = calloc(Z, sizeof *arr[i][k]);
Notes:
Here we assign values to the Y pointers that arr[i] points to, and this loop is inside the loop on i that executes X times, so this code assigns XY pointers in total.
So the answer to your question is we have 1 + X + XY pointers.
Nobody producing good commercial code uses this. Using pointers-to-pointers-to-pointers is bad for the hardware (meaning inefficient in performance) because the processor generally cannot predict where a pointer points to until it fetches it. Accessing some member of your array, arr[i][j][k], requires loading three pointers from memory.
In most C implementations, you can simply allocate a three-dimensional array:
double (*arr)[Y][Z] = calloc(X, sizeof *arr);
With this, when you access arr[i][j][k], the compiler will calculate the address (as, in effect, arr + (i*Y + j)*Z + k). Although that involves several multiplications and additions, they are fairly simple for modern processors and are likely as fast or faster than fetching pointers from memory and they leave the processor’s load-store unit free to fetch the actual array data. Also, when you are using the same i and/or j repeatedly, the compiler likely generates code that keeps i*Y and/or (i*Y + j)*Z around for multiple uses without recalculating them.
Well, short answer is: it is not known.
As a classic example, keep in mind the main() prototype
int main( int argc, char** argv);
argc keeps the number of pointers. Without it we do not know how many they are. The system builds the array argv, gently updates argc with the value and then launches the program.
Back to your array
double*** arr;
All you know is that
arr is a pointer.
*arr is double**, also a pointer
**arr is double*, also a pointer
***arr is a double.
What you will get in code depends on how you build this. A common way if you need an array of arrays and things like that is to mimic the system and use a few unsigned and wrap them all with the pointers into a struct like
typedef struct
{
int n_planes;
int n_rows;
int n_columns;
double*** array;
} Cube;
A CSV file for example is char ** **, a sheet workbook is char ** ** ** and it is a bit scary, but works. For each ** a counter is needed, as said above about main()
A C example
The code below uses arr, declared as double***, to
store a pointer to a pointer to a pointer to a double
prints the value using the 3 pointers
then uses arr again to build a cube of X*Y*Z doubles, using a bit of math to set values to 9XY9.Z9
the program uses 2, 3 and 4 for a total of 24 values
lists the full array
list the first and the very last element, arr[0][0][0] and arr[X-1][Y-1][Z-1]
frees the whole thing in reverse order
The code
#include <stdio.h>
#include <stdlib.h>
typedef struct
{
int n_planes;
int n_rows;
int n_columns;
double*** array;
} Cube;
int print_array(double***, int, int, int);
int main(void)
{
double sample = 20.21;
double* pDouble = &sample;
double** ppDouble = &pDouble;
double*** arr = &ppDouble;
printf("***arr is %.2ff\n", ***arr);
printf("original double is %.2ff\n", sample);
printf("*pDouble is %.2ff\n", *pDouble);
printf("**ppDouble is %.2ff\n", **ppDouble);
// but we can build a cube of XxYxZ doubles for arr
int X = 2;
int Y = 3;
int Z = 4; // 24 elements
arr = (double***)malloc(X * sizeof(double**));
// now each arr[i] must point to an array of double**
for (int i = 0; i < X; i += 1)
{
arr[i] = (double**)malloc(Y * sizeof(double*));
for (int j = 0; j < Y; j += 1)
{
arr[i][j] = (double*)malloc(Z * sizeof(double));
for (int k = 0; k < Z; k += 1)
{
arr[i][j][k] = (100. * i) + (10. * j) + (.1 * k) + 9009.09;
}
}
}
print_array(arr, X, Y, Z);
printf("\n\
Test: first element is arr[%d][%d[%d] = %6.2f (9XY9.Z9)\n\
last element is arr[%d][%d[%d] = %6.2f (9XY9.Z9)\n",
0, 0, 0, arr[0][0][0],
(X-1), (Y-1), (Z-1), arr[X-1][Y-1][Z-1]
);
// now to free this monster
for (int x = 0; x < X; x += 1)
{
for (int y = 0; y < Y; y += 1)
{
free(arr[x][y]); // the Z rows
}
free(arr[x]); // the plane Y
}
free(arr); // the initial pointer;
return 0;
}; // main()
int print_array(double*** block, int I, int J, int K)
{
for (int a = 0; a < I; a += 1)
{
printf("\nPlane %d\n\n", a);
for (int b = 0; b < J; b += 1)
{
for (int c = 0; c < K; c += 1)
{
printf("%6.2f ", block[a][b][c]);
}
printf("\n");
}
}
return 0;
}; // print_array()
The output
***arr is 20.21f
original double is 20.21f
*pDouble is 20.21f
**ppDouble is 20.21f
Plane 0
9009.09 9009.19 9009.29 9009.39
9019.09 9019.19 9019.29 9019.39
9029.09 9029.19 9029.29 9029.39
Plane 1
9109.09 9109.19 9109.29 9109.39
9119.09 9119.19 9119.29 9119.39
9129.09 9129.19 9129.29 9129.39
Test: first element is arr[0][0[0] = 9009.09 (9XY9.Z9)
last element is arr[1][2[3] = 9129.39 (9XY9.Z9)
I use nested data structure for fibonacci, but I have a segmentation fault 11.
void fib(int **fib_array, int n){
fib_array = malloc(n * sizeof(int*));
for(int i = 0; i < n; i++){
fib_array[i] = malloc(sizeof(int));
}
for(int i = 0; i < n; i++){
if (i <= 1){
fib_array[i][0] = i;
}
else{
fib_array[i][0] = fib_array[i - 2][0] + fib_array[i - 1][0];
}
}
}
int main(int argc, char **argv) {
/* do not change this main function */
int count = strtol(argv[1], NULL, 10);
int *fib_sequence;
fib(&fib_sequence, count);
for (int i = 0; i < count; i++) {
printf("%d ", fib_sequence[i]);
}
free(fib_sequence);
return 0;
}
you are being too complicated. You just need a single malloc
*fib_array = malloc(n * sizeof(int));
and remove you second indexings [0] from everywhere
The consfusion comes from **int. This looks like a multi dim array. Its not - its declared ** so that you can set the value in the caller. A simpler exampe will help
void Make42(int* v)
{
*v = 42;
}
int main()
{
int myv = 0;
Make42(&myv);
// now myv == 42
}
The * in the arg list is so that Make42 can 'reach out' and modify what was passed to it (myv in this case)
In your code the ** on fib array is there for the same purpose. you could have done (In know you werent allowed to by the test definition )
int *fib(int n){
int *fib_array = malloc(n * sizeof(int));
......
return fib_array;
}
and in main
fib_sequence = fib(count);
this makes it much clearer that you are really manipulating a simple array
pm100 is right, but a little short for answering to a beginner...
At first, you have passed a pointer to a pointer. If you want the original pointer to contain a value, you need to dereference the pointer to pointer:
*fib_array = ...
By assigning to the pointer only (as you did in your code), you do not modify the orignial pointer (fib_sequence in main) at all. And as you have not initialised it, it might point to anywhere, thus the segmentation fault when you try to print the values of it.
Then why an array of pointers to individually stored values? You can use a contiguous array of ints, which you get by
*fib_array = malloc(n * sizeof(int));
OK, further usage won't be too nice ((*fib_array)[i] = ...), so I recommend a temporary variable instead:
int* fa = malloc(n * sizeof(int));
// now fill in the values comfortably:
fa[i] = ...;
// finally, assign the pointer to the target:
*fib_array = fa;
Side note: always check the result of malloc, it could be NULL:
fa = ...
if(fa)
// assign values
else
// appropriate error handling
In your concrete case, you could omit the else branch in your function and check your pointer outside within main function.
By the way, a simple return value would have made your live easier, too:
int* fib(int n)
{
int* fib_array = malloc(n * sizeof(int*));
// ...
return fib_array;
}
Notice: no need for pointer to pointer... Usage:
int* fib_sequence = fib(count);
I'm rewriting a bunch of C99 VLAs for an open source project for support in Visual Studio. So I have a lot of statements like
void somefunc(double var[r]){...}
which I rewrite to
void somefunc(double *var) {
var = malloc(sizeof(double) * r);
...
free(var);
}
Which I am assuming to be the correct way to do this.
The problem is I am starting to see some more complicated VLA declarations which are leaving me stumped. For example:
double *(*var)[length1][length2][length3] = malloc(sizeof(double *[length4][length1][length2][length3]));
So to me this essentially looks like a 5 dimensional array.
This is clearly illegal in C89 because of the use of variables in the sizeof() function.
My guess on how to rewrite this would be:
double *****var = malloc(sizeof(double ****) * length1);
for(int i = 0; i<length1; i++) {
var[i] = malloc(sizeof(double ***) * length2);
for(int j = 0; j<length2; j++) {
var[i][j] = malloc(sizeof(double **) * length3);
for(int k = 0; k<length3; k++) {
var[i][j][k] = malloc(sizeof(double *) * length4);
}
}
}
As messy as it is, this is my best guess on how to rewrite the above statement. Is this technically correct, and is there a better way I should be going about this?
Here's my attempt:
double***** var = malloc(sizeof *var * length4);
for (int i = 0; i < length4; ++i)
{
var[i] = malloc(sizeof *var[i] * length1);
for (int j = 0; j < length1; ++j)
{
var[i][j] = malloc(sizeof *var[i][j] * length2);
for (int k = 0; k < length2; ++k)
{
var[i][j][k] = malloc(sizeof *var[i][j][k] * length3);
for (int l = 0; l < length3; ++l)
{
var[i][j][k][l] = NULL;
// var[i][j][k][l] is a double* - you can allocate some room for a double here, or assign the address of an existing variable
}
}
}
}
Similar to yours, except that note how I used length4 in the first malloc. If you look at your original declaration, var is a pointer to a three dimensional array of double*. As we can deduce from the malloc call on the right, memory for length4 such arrays is allocated, so you can think of var as an array of length4 three-dimensional arrays of double*.
I still maintain that anyone who puts such stuff in production code ought to be shot on the spot (well, you, as a maintainer, are excused).
IIRC declaring variables inside fors isn't valid C89, but you can move these above to the top of your scope.
Note that, as #immibis points out in the comments, your first conversion is most likely a mistake; double *var is an argument to the function, which means it's passed by value, so any changes you make to it inside the function aren't visible outside, and on top of that, you malloc some memory, do stuff with it, then free it. Even if you were modifying a passed pointer - which would require the parameter type to be double** - it still isn't necessary to pass a variable so you can use it locally exclusively. You most likely wanted to malloc the memory outside of the function and pass a valid pointer (and hopefully the size) to it.
I don't understand why this works:
void main() {
int * b;
b = (int *)malloc(sizeof(int));
*b = 1;
printf("*b = %d\n", *b);
}
while this does not (gets segmentation fault for the malloc()):
void main() {
int ** a;
int i;
for (i = 0; i<= 3; i++) {
a[i] = (int*)malloc(sizeof(int));
*(a[i]) = i;
printf("*a[%d] = %d\n", i, *(a[i]));
}
}
since I find a[i] is just like b in the first example.
BTW, a[i] is equal to *(a+i), right?
You need to allocate memory for a first, so that you can access its members as a[i].
So if you want to allocate for 4 int * do
a = malloc(sizeof(int *) * 4);
for (i = 0; i<= 3; i++) {
...
}
or define it as array of integer pointers as
int *a[4];
a is a 2 dimensional pointer, you have to allocate both dimension.
b is a 1 dimensional pointer, you have to allocate only one dimension and that's what you're doing with
b = (int *)malloc(sizeof(int));
So in order the second example to work you have to allocate the space for the pointer of pointer
void main() {
int ** a;
int i;
a = (int**)malloc(4*sizeof(int*));
for (i = 0; i<= 3; i++) {
a[i] = (int*)malloc(sizeof(int));
*(a[i]) = i;
printf("*a[%d] = %d\n", i, *(a[i]));
}
The allocated pointer is written to uninitialized memory (you never set a to anything), causing undefined behavior.
So no, it's not at all equivalent to the code in the first example.
You would need something like:
int **a;
a = malloc(3 * sizeof *a);
first, to make sure a holds something valid, then you can use indexing and assign to a[0].
Further, this:
a[i] = (int*)malloc(sizeof(int));
doesn't make any sense. It's assigning to a[i], an object of type int *, but allocating space for sizeof (int).
Finally, don't cast the return value of malloc() in C.
actually malloc it's not that trivial if you really want safe and portable, on linux for example malloc could return a positive response for a given request even if the actual memory it's not even really reserved for your program or the memory it's not writable.
For what I know both of your examples can potentially return a seg-fault or simply crash.
#ruppells-vulture I would argue that malloc is really portable and "safe" for this reasons.
I want to declare the correct pointer and allocate memory for a two-dimensional array, and correctly pass to a function. I am having trouble getting this to work properly and need help.
Here is my code:
double **podrucje;
podrucje=(double **) malloc (sizeof (double *) *123);
for (i=0;i<(123);i++)
{
podrucje[i]=(double *) malloc (sizeof (double) * 11);
}
for (i=0;i<(123);i++)
{
memset (podrucje[i], 0, 10);
}
But this code doesnt work, it messes up whole my program. So i decided to give up on dynamic allocation and use this:
double podrucje[123][11]={0};
But i dont know how to send it and use it in function...
memset works on per byte basis.
double **podrucje = (double **) malloc (sizeof (double *) * 123);
for (i = 0; i < 123; i++)
{
podrucje[i] = (double *) malloc (sizeof (double) * 11);
memset(podrucje[i], 0, sizeof(double) * 11);
}
if you want to pass it just declare it as such
void function(double podrucje[123][11]) {
...
}
You're best off to use malloc, but allocate the whole array on your second line, so it all gets allocated in contiguous memory. So
podrucje = (double*) malloc (sizeof (double) * 123 * 11);
Then the first loop can go away too. And it looks like you're initializing the array to 0 -- in that case, use calloc instead of malloc, eliminating the second loop.
To index into the array, use things like
double myitem = podrucje [11 * row + col];
You should of course use a define or similar to keep the use of 11 consistent, but that's not the point of this answer.
Write the function argument the same way you wrote the variable definition:
void myfunc(double podrucje[123][11])
{
...
}
double podrucje[123][11];
myfunc(podrucje);
Note that the array is passed "by reference" rather than being copied.
In the following code snippet,
podrucje[i]=(double *) malloc (sizeof (double) * 11);
for (i=0;i<(123);i++)
{
memset (podrucje[i], 0, 10);
}
1) You dont need the extra parenthesis against the numbers 123 and 11
The for loop can be as follows,
for (i = 0; i < 123; i++)
2) Instead of using 123 and 11. Try to define a MACRO and use that instead.
Advantage: The code becomes independent of special numbers and is easily maintainable. Especially in the cases of larger code base.
3) If you read the code, podrucje[i] is allocated a memory of 11 doubles But when you memset you are setting it for only 10 doubles while the last one may or may not consist of garbage. Use calloc here, It not only allocated memory but also initializes the same.
podrucje[i]=(double *) calloc(11,sizeof(double));
Also It could be more helpful if you could tell How exactly is it screwing up your code ?
Example, Code Snippet could help more than just stating its screwing up.
It helps others to investigate and solve the issue.
If you have a modern C compiler (C99 would do) you can even declare real 2D matrices with variable sizes. You don't need to fall back to this awful emulation with pointers.
void myfunc(size_t n, size_t m, double podrucje[n][m])
{
...
}
double (*podrucje)[n] = malloc(sizeof(double[n][m]));
myfunc(n, m, podrucje);
For the function you just have to ensure that the declarations of n and m come first, before the matrix.
There are several ways to dynamically allocate memory for an NxM array. Here are two:
You can declare a pointer to an M-element array, and then malloc N instances of it:
double (*podrucje)[11] = malloc(sizeof *podrucje * 123);
As of C89, you don't need to cast the result of malloc, and the practice is discouraged. Also, note that the operand to sizeof is the expression *podrucje; this gives me the same result as sizeof (double) * 11.
You would index this array as podrucje[i][j] like any other 2D array. podrucje[i] implicitly dereferences the pointer (remember that a[i] is equivalent to *(a + i)) so you don't have to do anything funky with it.
You would use it in a function as follows:
void init(double (*podrucje)[11], size_t rows)
{
size_t i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < 11; j++)
podrucje[i][j] = 0.0;
}
which would be called as
init(podrucje, 123);
The drawback to this method is that the function can only operate on Nx11 arrays; if you're using a C99 compiler or a C2011 compiler that supports variable length arrays, you could specify the number of columns as a runtime variable:
void foo(void)
{
size_t rows = 123, cols = 11;
double (*podrucje)[cols] = malloc(sizeof *podrucje * rows);
if (podrucje)
init(cols, podrucje, rows);
...
}
// cols must be declared before it can be used
// in an array declarator
//
void init(size_t cols, double(*podrucje)[cols], size_t rows)
{
size_t i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < cols; j++)
podrucje[i][j] = 0.0;
}
When you're done with the array, deallocate it as follows:
free(podrucje);
The other approach is to allocate each row separately, as follows:
size_t rows = 123, cols = 11;
double **podrucje = malloc(sizeof *podrucje * rows);
if (!podrucje)
{
// malloc failed; handle allocation error here
}
else
{
size_t i;
for (i = 0; i < rows; i++)
{
podrucje[i] = malloc(sizeof *podrucje[i] * cols);
if (!podrucje[i])
{
// malloc failed; handle allocation error here
}
}
}
And you would use it in a function as follows:
void foo()
{
double **podrucje;
// allocate array as above
init(foo, rows, cols);
...
}
void init(double **podrucje, size_t rows, size_t cols)
{
size_t i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < cols; j++)
podrucje[i][j] = 0.0;
}
When you're finished with the array, deallocate it as follows:
for(i = 0; i < rows; i++)
free(podrucje[i]);
free(podrucje);
The first method allocates memory as a single, contiguous block; the second allocates it in a series of smaller, discontinuous chunks. If your array is especially big or your heap especially fragmented, the first method may fail where the second will succeed. If you're working with a compiler that doesn't support variable-length arrays, the first method is much less flexible, because the number of columns must be specified at compile time.
How could the same indexing method work for both forms?
In the first case, each podrucje[i] is an 11-element array of double; indexing it with j works like any other array. In the second case, each podrucje[i] is a pointer to double. Since a[i] is evaluated as *(a + i), array indexing works on pointer expressions just as well as array expressions.
int print_func(char((*p)[26])[10])
{
int i = 0;
for (i=0; i < 26 ; i++) {
fprintf(stderr, "%02d:%.*s\n", i, 10, p[0][i]);
}
return (0);
}
int main(void)
{
int nrow = 26;
int ncol = 10;
char((*p)[26])[10] = (char((*)[26])[10])(0);
char(*q)[10];
char c = 'a';
int i = 0;
p = (char((*)[26])[10])malloc(sizeof(char) * nrow * ncol);
if ((char((*)[26])[10])0 == p) {
return (-1);
}
for (i=0, q=p[0]; i < nrow ; i++) {
memset(q, c, sizeof(char) * ncol);
c++;
q++;
}
for (i=0,q=p[0] ; i < nrow ; i++) {
fprintf(stderr, "%.*s\n", 10, q);
q++;
}
p[0][8][0]='z';
getchar();
print_func(p);
return (0);
}