Related
I am somewhat new to C programming. I have a doubt regarding dynamic memory allocation. The following is a code in the main program for memory allocation.
double **mat=(double**)malloc(sizeof(double*)*n);
mat[0]=(double*)calloc(sizeof(double),n*n);
for(i=1; i<n; i++)
mat[i] = mat[i-1] + n;
mat = create_square_matrix(n);
I want to call the function and create elements in the matrix inside the function. Do I have once again allocation memory inside the function like below or Is there any other method to avoid this tedious memory allocation repetition. Following is the function.
`double** create_square_matrix(int n)
{
int i,j,sum=0;
double **array2=(double**)malloc(sizeof(double*)*n);
array2[0]=(double*)calloc(sizeof(double),n*n);
for(i=1; i<n; i++)
array2[i] = array2[i-1] + n;
for (i = 0; i < n; ++i)
{
for (j = 0; j < n; ++j)
{
sum=sum+j;
array2[i][j]=sum;
}
}
return array2;
}
`
The above function returns the array which is stored in the 'mat' variable. And another question is how do I free the memory of variable 'array2' inside the function after using the return method. I can't free the memory possibly before returning the array. Is there a method to free the memory in the above function.
Your function create_square_matrix allocates memory and then fills it by some values.
Your top piece of code allocates memory, and then calls create_square_matrix which again allocates memory. It is like to mop floors before calling the janitor who also mops floors. You don't need to allocate memory twice. Not only it is unneccessary, in fact it is bad. Since you perform two allocations, the memory from the first one is lost, and there is no way to free it. This is called memory leak. Instead of
double **mat=(double**)malloc(sizeof(double*)*n);
mat[0]=(double*)calloc(sizeof(double),n*n);
for(i=1; i<n; i++)
mat[i] = mat[i-1] + n;
mat = create_square_matrix(n);
you should write
double **mat = create_square_matrix(n);
As mentioned, in C there's no need to cast to/from void*. Also, your calloc is backwards: use N elements for the first argument, sizeof(element) for the second.
I will answer your question, and then suggest a better approach.
You are choosing double ** for your matrix. Even if it weren't square, you know exactly how many pointers and how many doubles you need, and of course how big each type is. So,
double **M = malloc( n * sizeof(double*) + n * m * sizeof(double) );
does the trick, does it not? If sizeof(double*) == sizeof(double) for your machine (probably true), then
double **M = calloc( (1+n) * m, sizeof(double) );
works, too, but less portably. You also get the zeros for free; you'll have trouble finding a machine for which it's not true that double f=0 yields a value for which all bits are zero.
But why define your matrix as an array of pointers? Why not instead define it as an array of doubles?
double *M = calloc( n * m, sizeof(double) );
Better yet, for the past 15 years or so, C has supported variable-length arrays, meaning you can define arrays whose size is determined at runtime. Back in K&R days, you could define an array M[n] unless n was a static constant or enum. If your arrays aren't ginormous -- meaning that for the machine in question they'll fit comfortably on the stack -- you can skip malloc and simply define your matrix by a size determined at runtime.
Even if you can't do that, can typedef one dimension dynamically,
typedef double (x_axis_t)[m];
double x_axis_t *M = calloc( n * sizeof(x_axis_t), sizeof(double) );
which is nice, because then you can access your array as
M[x][y];
Just don't try to use M[x,y], because that's something else altogether.
BTW, since you're new to the game, to use a c99 compiler, the standard command is c99, not cc. See your friendly manual for details. :-)
Using a nice function macro for memory allocation is always a good idea. Unless you have to free memory manually I would leave it to a garbage collector such as libgc. Below is an example. If you don't want to use a garbage collector you can just replace GC_MALLOC with malloc. When you free the array (manually) you must first free the individual rows.
#include <gc/gc.h>
#include <stdio.h>
#include <stdlib.h>
#define NEW_ARRAY(ptr, n) (ptr) = GC_MALLOC((n) * sizeof (ptr)[0])
double **SquareMatrix(int n)
{
double **A;
int i, j;
NEW_ARRAY(A, n);
for (i = 0; i < n; i++) {
NEW_ARRAY(A[i], n);
for (j = 0; j < n; j++) {
A[i][j] = 0.0;
}
}
return A;
}
int main(void)
{
const int n = 5;
double **A;
int i, j;
A = SquareMatrix(n);
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
printf("%5.2f ", A[i][j]);
}
putchar('\n');
}
return 0;
}
Does someone know how I can use dynamically allocated multi-dimensional arrays using C? Is that possible?
Since C99, C has 2D arrays with dynamical bounds. If you want to avoid that such beast are allocated on the stack (which you should), you can allocate them easily in one go as the following
double (*A)[n] = malloc(sizeof(double[n][n]));
and that's it. You can then easily use it as you are used for 2D arrays with something like A[i][j]. And don't forget that one at the end
free(A);
Randy Meyers wrote series of articles explaining variable length arrays (VLAs).
With dynamic allocation, using malloc:
int** x;
x = malloc(dimension1_max * sizeof(*x));
for (int i = 0; i < dimension1_max; i++) {
x[i] = malloc(dimension2_max * sizeof(x[0]));
}
//Writing values
x[0..(dimension1_max-1)][0..(dimension2_max-1)] = Value;
[...]
for (int i = 0; i < dimension1_max; i++) {
free(x[i]);
}
free(x);
This allocates an 2D array of size dimension1_max * dimension2_max. So, for example, if you want a 640*480 array (f.e. pixels of an image), use dimension1_max = 640, dimension2_max = 480. You can then access the array using x[d1][d2] where d1 = 0..639, d2 = 0..479.
But a search on SO or Google also reveals other possibilities, for example in this SO question
Note that your array won't allocate a contiguous region of memory (640*480 bytes) in that case which could give problems with functions that assume this. So to get the array satisfy the condition, replace the malloc block above with this:
int** x;
int* temp;
x = malloc(dimension1_max * sizeof(*x));
temp = malloc(dimension1_max * dimension2_max * sizeof(x[0]));
for (int i = 0; i < dimension1_max; i++) {
x[i] = temp + (i * dimension2_max);
}
[...]
free(temp);
free(x);
Basics
Arrays in c are declared and accessed using the [] operator. So that
int ary1[5];
declares an array of 5 integers. Elements are numbered from zero so ary1[0] is the first element, and ary1[4] is the last element. Note1: There is no default initialization, so the memory occupied by the array may initially contain anything. Note2: ary1[5] accesses memory in an undefined state (which may not even be accessible to you), so don't do it!
Multi-dimensional arrays are implemented as an array of arrays (of arrays (of ... ) ). So
float ary2[3][5];
declares an array of 3 one-dimensional arrays of 5 floating point numbers each. Now ary2[0][0] is the first element of the first array, ary2[0][4] is the last element of the first array, and ary2[2][4] is the last element of the last array. The '89 standard requires this data to be contiguous (sec. A8.6.2 on page 216 of my K&R 2nd. ed.) but seems to be agnostic on padding.
Trying to go dynamic in more than one dimension
If you don't know the size of the array at compile time, you'll want to dynamically allocate the array. It is tempting to try
double *buf3;
buf3 = malloc(3*5*sizeof(double));
/* error checking goes here */
which should work if the compiler does not pad the allocation (stick extra space between the one-dimensional arrays). It might be safer to go with:
double *buf4;
buf4 = malloc(sizeof(double[3][5]));
/* error checking */
but either way the trick comes at dereferencing time. You can't write buf[i][j] because buf has the wrong type. Nor can you use
double **hdl4 = (double**)buf;
hdl4[2][3] = 0; /* Wrong! */
because the compiler expects hdl4 to be the address of an address of a double. Nor can you use double incomplete_ary4[][]; because this is an error;
So what can you do?
Do the row and column arithmetic yourself
Allocate and do the work in a function
Use an array of pointers (the mechanism qrdl is talking about)
Do the math yourself
Simply compute memory offset to each element like this:
for (i=0; i<3; ++i){
for(j=0; j<3; ++j){
buf3[i * 5 + j] = someValue(i,j); /* Don't need to worry about
padding in this case */
}
}
Allocate and do the work in a function
Define a function that takes the needed size as an argument and proceed as normal
void dary(int x, int y){
double ary4[x][y];
ary4[2][3] = 5;
}
Of course, in this case ary4 is a local variable and you can not return it: all the work with the array must be done in the function you call of in functions that it calls.
An array of pointers
Consider this:
double **hdl5 = malloc(3*sizeof(double*));
/* Error checking */
for (i=0; i<3; ++i){
hdl5[i] = malloc(5*sizeof(double))
/* Error checking */
}
Now hdl5 points to an array of pointers each of which points to an array of doubles. The cool bit is that you can use the two-dimensional array notation to access this structure---hdl5[0][2] gets the middle element of the first row---but this is none-the-less a different kind of object than a two-dimensional array declared by double ary[3][5];.
This structure is more flexible then a two dimensional array (because the rows need not be the same length), but accessing it will generally be slower and it requires more memory (you need a place to hold the intermediate pointers).
Note that since I haven't setup any guards you'll have to keep track of the size of all the arrays yourself.
Arithmetic
c provides no support for vector, matrix or tensor math, you'll have to implement it yourself, or bring in a library.
Multiplication by a scaler and addition and subtraction of arrays of the same rank are easy: just loop over the elements and perform the operation as you go. Inner products are similarly straight forward.
Outer products mean more loops.
If you know the number of columns at compile time, it's pretty simple:
#define COLS ...
...
size_t rows;
// get number of rows
T (*ap)[COLS] = malloc(sizeof *ap * rows); // ap is a *pointer to an array* of T
You can treat ap like any 2D array:
ap[i][j] = x;
When you're done you deallocate it as
free(ap);
If you don't know the number of columns at compile time, but you're working with a C99 compiler or a C2011 compiler that supports variable-length arrays, it's still pretty simple:
size_t rows;
size_t cols;
// get rows and cols
T (*ap)[cols] = malloc(sizeof *ap * rows);
...
ap[i][j] = x;
...
free(ap);
If you don't know the number of columns at compile time and you're working with a version of C that doesn't support variable-length arrays, then you'll need to do something different. If you need all of the elements to be allocated in a contiguous chunk (like a regular array), then you can allocate the memory as a 1D array, and compute a 1D offset:
size_t rows, cols;
// get rows and columns
T *ap = malloc(sizeof *ap * rows * cols);
...
ap[i * rows + j] = x;
...
free(ap);
If you don't need the memory to be contiguous, you can follow a two-step allocation method:
size_t rows, cols;
// get rows and cols
T **ap = malloc(sizeof *ap * rows);
if (ap)
{
size_t i = 0;
for (i = 0; i < cols; i++)
{
ap[i] = malloc(sizeof *ap[i] * cols);
}
}
ap[i][j] = x;
Since allocation was a two-step process, deallocation also needs to be a two-step process:
for (i = 0; i < cols; i++)
free(ap[i]);
free(ap);
malloc will do.
int rows = 20;
int cols = 20;
int *array;
array = malloc(rows * cols * sizeof(int));
Refer the below article for help:-
http://courses.cs.vt.edu/~cs2704/spring00/mcquain/Notes/4up/Managing2DArrays.pdf
Here is working code that defines a subroutine make_3d_array to allocate a multidimensional 3D array with N1, N2 and N3 elements in each dimension, and then populates it with random numbers. You can use the notation A[i][j][k] to access its elements.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
// Method to allocate a 2D array of floats
float*** make_3d_array(int nx, int ny, int nz) {
float*** arr;
int i,j;
arr = (float ***) malloc(nx*sizeof(float**));
for (i = 0; i < nx; i++) {
arr[i] = (float **) malloc(ny*sizeof(float*));
for(j = 0; j < ny; j++) {
arr[i][j] = (float *) malloc(nz * sizeof(float));
}
}
return arr;
}
int main(int argc, char *argv[])
{
int i, j, k;
size_t N1=10,N2=20,N3=5;
// allocates 3D array
float ***ran = make_3d_array(N1, N2, N3);
// initialize pseudo-random number generator
srand(time(NULL));
// populates the array with random numbers
for (i = 0; i < N1; i++){
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
ran[i][j][k] = ((float)rand()/(float)(RAND_MAX));
}
}
}
// prints values
for (i=0; i<N1; i++) {
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
printf("A[%d][%d][%d] = %f \n", i,j,k,ran[i][j][k]);
}
}
}
free(ran);
}
There's no way to allocate the whole thing in one go. Instead, create an array of pointers, then, for each pointer, create the memory for it. For example:
int** array;
array = (int**)malloc(sizeof(int*) * 50);
for(int i = 0; i < 50; i++)
array[i] = (int*)malloc(sizeof(int) * 50);
Of course, you can also declare the array as int* array[50] and skip the first malloc, but the second set is needed in order to dynamically allocate the required storage.
It is possible to hack a way to allocate it in a single step, but it would require a custom lookup function, but writing that in such a way that it will always work can be annoying. An example could be L(arr,x,y,max_x) arr[(y)*(max_x) + (x)], then malloc a block of 50*50 ints or whatever and access using that L macro, e.g.
#define L(arr,x,y,max_x) arr[(y)*(max_x) + (x)]
int dim_x = 50;
int dim_y = 50;
int* array = malloc(dim_x*dim_y*sizeof(int));
int foo = L(array, 4, 6, dim_x);
But that's much nastier unless you know the effects of what you're doing with the preprocessor macro.
int rows, columns;
/* initialize rows and columns to the desired value */
arr = (int**)malloc(rows*sizeof(int*));
for(i=0;i<rows;i++)
{
arr[i] = (int*)malloc(cols*sizeof(int));
}
// use new instead of malloc as using malloc leads to memory leaks
`enter code here
int **adj_list = new int*[rowsize];
for(int i = 0; i < rowsize; ++i)
{
adj_list[i] = new int[colsize];
}
How to allocate dynamic memory for 2d array in function ?
I tried this way:
int main()
{
int m=4,n=3;
int** arr;
allocate_mem(&arr,n,m);
}
void allocate_mem(int*** arr,int n, int m)
{
*arr=(int**)malloc(n*sizeof(int*));
for(int i=0;i<n;i++)
*arr[i]=(int*)malloc(m*sizeof(int));
}
But it doesn't work.
Your code is wrong at *arr[i]=(int*)malloc(m*sizeof(int)); because the precedence of the [] operator is higher than the * deference operator: In the expression *arr[i], first arr[i] is evaluated then * is applied. What you need is the reverse (dereference arr, then apply []).
Use parentheses like this: (*arr)[i] to override operator precedence. Now, your code should look like this:
void allocate_mem(int*** arr, int n, int m)
{
*arr = (int**)malloc(n*sizeof(int*));
for(int i=0; i<n; i++)
(*arr)[i] = (int*)malloc(m*sizeof(int));
}
To understand further what happens in the above code, read this answer.
It is important that you always deallocate dynamically allocated memory explicitly once you are done working with it. To free the memory allocated by the above function, you should do this:
void deallocate_mem(int*** arr, int n){
for (int i = 0; i < n; i++)
free((*arr)[i]);
free(*arr);
}
Additionally, a better way to create a 2D array is to allocate contiguous memory with a single malloc() function call as below:
int* allocate_mem(int*** arr, int n, int m)
{
*arr = (int**)malloc(n * sizeof(int*));
int *arr_data = malloc( n * m * sizeof(int));
for(int i=0; i<n; i++)
(*arr)[i] = arr_data + i * m ;
return arr_data; //free point
}
To deallocate this memory:
void deallocate_mem(int*** arr, int* arr_data){
free(arr_data);
free(*arr);
}
Notice that in the second technique malloc is called only two times, and so in the deallocation code free is called only two times instead of calling it in a loop. So this technique should be better.
Consider this: Just single allocation
int** allocate2D(int m, int n)
{
int **a = (int **)malloc(m * sizeof(int *) + (m * n * sizeof(int)));
int *mem = (int *)(a + m);
for(int i = 0; i < m; i++)
{
a[i] = mem + (i * n);
}
return a;
}
To Free:
free(a);
If your array does not need to be resized (well, you can, but il will be a bit more complicated), there is an easier/more efficient way to build 2D arrays in C.
Take a look at http://c-faq.com/aryptr/dynmuldimary.html.
The second method (for the array called array2) is quite simple, less painful (try to add the tests for mallocs' return value), and way more efficient.
I've just benchmarked it, for a 200x100 array, allocated and deallocated 100000 times:
Method 1 : 1.8s
Method 2 : 47ms
And the data in the array will be more contiguous, which may speed things up (you may get some more efficient techniques to copy, reset... an array allocated this way).
Rather allocating the memory in many different block, one can allocate this in a consecutive block of memory.
Do the following:
int** my2DAllocation(int rows,int columns)
{
int i;
int header= rows *sizeof(int *);
int data=rows*cols*sizeof(int);
int ** rowptr=(int **)malloc(header+data);
if(rowptr==NULL)
{
return NULL:
}
int * buf=(int*)(rowptr+rows);
for(i=0;i<rows;i++)
{
rowptr[i]=buf+i*cols;
}
return rowptr;
}
That is an unnecessarily complicated way of allocating space for an array. Consider this idiom:
int main(void) {
size_t m = 4, n = 3;
int (*array)[m];
array = malloc(n * sizeof *array);
free(array);
}
I have tried the following code for allocating memory to 2 dimensional array.
#include<stdio.h>
#include<malloc.h>
void main(void)
{
int **p;//double pointer holding a 2d array
int i,j;
for(i=0;i<3;i++)
{
p=(int**)(malloc(sizeof(int*)));//memory allocation for double pointer
for(j=(3*i+1);j<(3*i+4);j++)
{
*p = (int*)(malloc(sizeof(int)));//memory allocation for pointer holding integer array
**p = j;
printf(" %d",**p);//print integers in a row
printf("\n");
p++;
}
}
}
Output of the above code is:-
1 2 3
4 5 6
7 8 9
In order to understand 2 dimensional array in terms of pointers, we need to understand how it will be allocated in memory, it should be something like this:-
1 2 3
1000 --> 100 104 108
4 5 6
1004 --> 200 204 208
7 8 9
1008 --> 300 304 308
from the above, we understand that, when we allocate memory to pointer p which is a double pointer, it is pointing to an array of integers, so in this example, we see that the 0x1000 is pointer p.
This pointer is pointing to integer pointer *p which is array of integers, when memory is allocated inside the inner for loop, during first iteration the pointer is 0x100 which is pointing to integer value 1, when we assign **p = j. Similarly it will be pointing to 2 and 3 in the next iterations in the loop.
Before the next iteration of the outer loop, double pointer is incremented, inside the next iteration, as is seen in this example the pointer is now at 0x1004 and is pointing to integer pointer which is an array of integers 4,5,6 and similarly for the next iterations in the loop.
Try the following code:
void allocate_mem(int*** arr,int n, int m)
{
*arr=(int**)malloc(n*sizeof(int*));
for(int i=0;i<n;i++)
*(arr+i)=(int*)malloc(m*sizeof(int));
}
2d Array dynamically array using malloc:
int row = 4;
int column = 4;
int val = 2;
// memory allocation using malloc
int **arrM = (int**)malloc (row*sizeof(int*));
for (int i=0;i<row;i++)
{
arrM[i] = (int*)malloc(column*sizeof(int));
// insert the value for each field
for (int j =0;j<column;j++,val++)
{
arrM[i][j] = val;
}
}
// De-allocation
for (int i=0;i<row;i++)
{
free(arrM[i]);
}
free(arrM);
arrM = 0;
//
// Now using New operator:
//
int **arr = new int*[row];
int k = 1;
for (int i=0;i<row;i++)
{
arr[i] = new int[column];
// insert the value for each field
for (int j =0;j<column;j++,k++)
{
arr[i][j] = k;
}
}
cout<<"array value is = "<<*(*(arr+0)+0)<<endl;
cout<<"array value is = "<<*(*(arr+3)+2)<<endl;
// Need to deallcate memory;
for (int i=0;i<row;i++)
{
delete [] arr[i];
}
delete []arr;
arr = 0;
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);
}
Does someone know how I can use dynamically allocated multi-dimensional arrays using C? Is that possible?
Since C99, C has 2D arrays with dynamical bounds. If you want to avoid that such beast are allocated on the stack (which you should), you can allocate them easily in one go as the following
double (*A)[n] = malloc(sizeof(double[n][n]));
and that's it. You can then easily use it as you are used for 2D arrays with something like A[i][j]. And don't forget that one at the end
free(A);
Randy Meyers wrote series of articles explaining variable length arrays (VLAs).
With dynamic allocation, using malloc:
int** x;
x = malloc(dimension1_max * sizeof(*x));
for (int i = 0; i < dimension1_max; i++) {
x[i] = malloc(dimension2_max * sizeof(x[0]));
}
//Writing values
x[0..(dimension1_max-1)][0..(dimension2_max-1)] = Value;
[...]
for (int i = 0; i < dimension1_max; i++) {
free(x[i]);
}
free(x);
This allocates an 2D array of size dimension1_max * dimension2_max. So, for example, if you want a 640*480 array (f.e. pixels of an image), use dimension1_max = 640, dimension2_max = 480. You can then access the array using x[d1][d2] where d1 = 0..639, d2 = 0..479.
But a search on SO or Google also reveals other possibilities, for example in this SO question
Note that your array won't allocate a contiguous region of memory (640*480 bytes) in that case which could give problems with functions that assume this. So to get the array satisfy the condition, replace the malloc block above with this:
int** x;
int* temp;
x = malloc(dimension1_max * sizeof(*x));
temp = malloc(dimension1_max * dimension2_max * sizeof(x[0]));
for (int i = 0; i < dimension1_max; i++) {
x[i] = temp + (i * dimension2_max);
}
[...]
free(temp);
free(x);
Basics
Arrays in c are declared and accessed using the [] operator. So that
int ary1[5];
declares an array of 5 integers. Elements are numbered from zero so ary1[0] is the first element, and ary1[4] is the last element. Note1: There is no default initialization, so the memory occupied by the array may initially contain anything. Note2: ary1[5] accesses memory in an undefined state (which may not even be accessible to you), so don't do it!
Multi-dimensional arrays are implemented as an array of arrays (of arrays (of ... ) ). So
float ary2[3][5];
declares an array of 3 one-dimensional arrays of 5 floating point numbers each. Now ary2[0][0] is the first element of the first array, ary2[0][4] is the last element of the first array, and ary2[2][4] is the last element of the last array. The '89 standard requires this data to be contiguous (sec. A8.6.2 on page 216 of my K&R 2nd. ed.) but seems to be agnostic on padding.
Trying to go dynamic in more than one dimension
If you don't know the size of the array at compile time, you'll want to dynamically allocate the array. It is tempting to try
double *buf3;
buf3 = malloc(3*5*sizeof(double));
/* error checking goes here */
which should work if the compiler does not pad the allocation (stick extra space between the one-dimensional arrays). It might be safer to go with:
double *buf4;
buf4 = malloc(sizeof(double[3][5]));
/* error checking */
but either way the trick comes at dereferencing time. You can't write buf[i][j] because buf has the wrong type. Nor can you use
double **hdl4 = (double**)buf;
hdl4[2][3] = 0; /* Wrong! */
because the compiler expects hdl4 to be the address of an address of a double. Nor can you use double incomplete_ary4[][]; because this is an error;
So what can you do?
Do the row and column arithmetic yourself
Allocate and do the work in a function
Use an array of pointers (the mechanism qrdl is talking about)
Do the math yourself
Simply compute memory offset to each element like this:
for (i=0; i<3; ++i){
for(j=0; j<3; ++j){
buf3[i * 5 + j] = someValue(i,j); /* Don't need to worry about
padding in this case */
}
}
Allocate and do the work in a function
Define a function that takes the needed size as an argument and proceed as normal
void dary(int x, int y){
double ary4[x][y];
ary4[2][3] = 5;
}
Of course, in this case ary4 is a local variable and you can not return it: all the work with the array must be done in the function you call of in functions that it calls.
An array of pointers
Consider this:
double **hdl5 = malloc(3*sizeof(double*));
/* Error checking */
for (i=0; i<3; ++i){
hdl5[i] = malloc(5*sizeof(double))
/* Error checking */
}
Now hdl5 points to an array of pointers each of which points to an array of doubles. The cool bit is that you can use the two-dimensional array notation to access this structure---hdl5[0][2] gets the middle element of the first row---but this is none-the-less a different kind of object than a two-dimensional array declared by double ary[3][5];.
This structure is more flexible then a two dimensional array (because the rows need not be the same length), but accessing it will generally be slower and it requires more memory (you need a place to hold the intermediate pointers).
Note that since I haven't setup any guards you'll have to keep track of the size of all the arrays yourself.
Arithmetic
c provides no support for vector, matrix or tensor math, you'll have to implement it yourself, or bring in a library.
Multiplication by a scaler and addition and subtraction of arrays of the same rank are easy: just loop over the elements and perform the operation as you go. Inner products are similarly straight forward.
Outer products mean more loops.
If you know the number of columns at compile time, it's pretty simple:
#define COLS ...
...
size_t rows;
// get number of rows
T (*ap)[COLS] = malloc(sizeof *ap * rows); // ap is a *pointer to an array* of T
You can treat ap like any 2D array:
ap[i][j] = x;
When you're done you deallocate it as
free(ap);
If you don't know the number of columns at compile time, but you're working with a C99 compiler or a C2011 compiler that supports variable-length arrays, it's still pretty simple:
size_t rows;
size_t cols;
// get rows and cols
T (*ap)[cols] = malloc(sizeof *ap * rows);
...
ap[i][j] = x;
...
free(ap);
If you don't know the number of columns at compile time and you're working with a version of C that doesn't support variable-length arrays, then you'll need to do something different. If you need all of the elements to be allocated in a contiguous chunk (like a regular array), then you can allocate the memory as a 1D array, and compute a 1D offset:
size_t rows, cols;
// get rows and columns
T *ap = malloc(sizeof *ap * rows * cols);
...
ap[i * rows + j] = x;
...
free(ap);
If you don't need the memory to be contiguous, you can follow a two-step allocation method:
size_t rows, cols;
// get rows and cols
T **ap = malloc(sizeof *ap * rows);
if (ap)
{
size_t i = 0;
for (i = 0; i < cols; i++)
{
ap[i] = malloc(sizeof *ap[i] * cols);
}
}
ap[i][j] = x;
Since allocation was a two-step process, deallocation also needs to be a two-step process:
for (i = 0; i < cols; i++)
free(ap[i]);
free(ap);
malloc will do.
int rows = 20;
int cols = 20;
int *array;
array = malloc(rows * cols * sizeof(int));
Refer the below article for help:-
http://courses.cs.vt.edu/~cs2704/spring00/mcquain/Notes/4up/Managing2DArrays.pdf
Here is working code that defines a subroutine make_3d_array to allocate a multidimensional 3D array with N1, N2 and N3 elements in each dimension, and then populates it with random numbers. You can use the notation A[i][j][k] to access its elements.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
// Method to allocate a 2D array of floats
float*** make_3d_array(int nx, int ny, int nz) {
float*** arr;
int i,j;
arr = (float ***) malloc(nx*sizeof(float**));
for (i = 0; i < nx; i++) {
arr[i] = (float **) malloc(ny*sizeof(float*));
for(j = 0; j < ny; j++) {
arr[i][j] = (float *) malloc(nz * sizeof(float));
}
}
return arr;
}
int main(int argc, char *argv[])
{
int i, j, k;
size_t N1=10,N2=20,N3=5;
// allocates 3D array
float ***ran = make_3d_array(N1, N2, N3);
// initialize pseudo-random number generator
srand(time(NULL));
// populates the array with random numbers
for (i = 0; i < N1; i++){
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
ran[i][j][k] = ((float)rand()/(float)(RAND_MAX));
}
}
}
// prints values
for (i=0; i<N1; i++) {
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
printf("A[%d][%d][%d] = %f \n", i,j,k,ran[i][j][k]);
}
}
}
free(ran);
}
There's no way to allocate the whole thing in one go. Instead, create an array of pointers, then, for each pointer, create the memory for it. For example:
int** array;
array = (int**)malloc(sizeof(int*) * 50);
for(int i = 0; i < 50; i++)
array[i] = (int*)malloc(sizeof(int) * 50);
Of course, you can also declare the array as int* array[50] and skip the first malloc, but the second set is needed in order to dynamically allocate the required storage.
It is possible to hack a way to allocate it in a single step, but it would require a custom lookup function, but writing that in such a way that it will always work can be annoying. An example could be L(arr,x,y,max_x) arr[(y)*(max_x) + (x)], then malloc a block of 50*50 ints or whatever and access using that L macro, e.g.
#define L(arr,x,y,max_x) arr[(y)*(max_x) + (x)]
int dim_x = 50;
int dim_y = 50;
int* array = malloc(dim_x*dim_y*sizeof(int));
int foo = L(array, 4, 6, dim_x);
But that's much nastier unless you know the effects of what you're doing with the preprocessor macro.
int rows, columns;
/* initialize rows and columns to the desired value */
arr = (int**)malloc(rows*sizeof(int*));
for(i=0;i<rows;i++)
{
arr[i] = (int*)malloc(cols*sizeof(int));
}
// use new instead of malloc as using malloc leads to memory leaks
`enter code here
int **adj_list = new int*[rowsize];
for(int i = 0; i < rowsize; ++i)
{
adj_list[i] = new int[colsize];
}