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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];
}
I have seen dozens of questions about “what’s wrong with my code” regarding multidimensional arrays in C. For some reason people can’t seem to wrap their head around what is happening here, so I decided to answer this question as a reference to others:
How do I correctly set up, access, and free a multidimensional array in C?
If others have helpful advice, please feel free to post along!
In C since C99, even dynamic multidimensional arrays can be easily allocated in one go with malloc and freed with free:
double (*A)[n] = malloc(sizeof(double[n][n]));
for (size_t i = 0; i < n; ++i)
for (size_t j = 0; j < n; ++j)
A[i][j] = someinvolvedfunction(i, j);
free(A);
There are at least four different ways to create or simulate a multi-dimensional array in C89.
One is "allocate each row separately", described by Mike in his answer. It is not a multidimensional array, it merely imitates one (in particular it mimics the syntax for accessing an element). It can be useful in the case where each row has different size, so you aren't representing a matrix but rather something with a "ragged edge".
One is "allocate a multidimensional array". It looks likes this:
int (*rows)[NUM_ROWS][NUM_COLS] = malloc(sizeof *rows);
...
free(rows);
Then the syntax to access element [i,j] is (*rows)[i][j]. In C89, both NUM_COLS and NUM_ROWS must be known at compile-time. This is a true 2-D array, and rows is a pointer to it.
One is, "allocate an array of rows". It looks like this:
int (*rows)[NUM_COLS] = malloc(sizeof(*rows) * NUM_ROWS);
...
free(rows);
Then the syntax to access element [i,j] is rows[i][j]. In C89, NUM_COLS must be known at compile-time. This is a true 2-D array.
One is, "allocate a 1-d array and pretend". It looks like this:
int *matrix = malloc(sizeof(int) * NUM_COLS * NUM_ROWS);
...
free(matrix);
Then the syntax to access element [i,j] is matrix[NUM_COLS * i + j]. This (of course) is not a true 2-D array. In practice it has the same layout as one.
Statically speaking, this is easy to understand:
int mtx[3][2] = {{1, 2},
{2, 3},
{3, 4}};
Nothing complicated here. 3 rows, 2 columns; data in column one: 1, 2, 3; data in column two: 2, 3, 4.
We can access the elements via the same construct:
for(i = 0; i<3; i++){
for(j = 0; j<2; j++)
printf("%d ", mtx[i][j]);
printf("\n");
}
//output
//1 2
//2 3
//3 4
Now let’s look at this in terms of Pointers:
The brackets are a very nice construct to help simplify things, but it doesn’t help when we need to work in a dynamic environment, so we need to think of this in terms of pointers. If we want to store a “row” of integers, we need an array:
int row[2] = {1,2};
And you know what? We can access this just like a pointer.
printf("%d, %d\n",*row,*(row+1)); //prints 1, 2
printf("%d, %d\n",row[0],row[1]); //prints 1, 2
Now if we don’t know the number of values in a row we can make this array a dynamic length if we have a pointer to int, and we give it some memory:
int *row = malloc(X * sizeof(int)); //allow for X number of ints
*row = 1; //row[0] = 1
*(row+1) = 2; //row[1] = 2
…
*(row+(X-1)) = Y; // row[x-1] = Some value y
So now we have a dynamic 1 dimensional array; a single row. But we want lots of rows, not just one, and we don’t know how many. That means we need another dynamic 1 dimensional array, each element of that array will be a pointer which points to a row.
//we want enough memory to point to X number of rows
//each value stored there is a pointer to an integer
int ** matrix = malloc(X * sizeof(int *));
//conceptually:
(ptr to ptr to int) (pointer to int)
**matrix ------------> *row1 --------> [1][2]
*row2 --------> [2][3]
*row3 --------> [3][4]
Now all that’s left to do is to write the code which will perform these dynamic allocations:
int i, j, value = 0;
//allocate memory for the pointers to rows
int ** matrix = malloc(Rows * sizeof(int*));
//each row needs a dynamic number of elements
for(i=0; i<Rows; i++){
// so we need memory for the number of items in each row…
// we could call this number of columns as well
*(matrix + i) = malloc(X * sizeof(int));
//While we’re in here, if we have the items we can populate the matrix
for(j=0; j<X; j++)
*(*(matrix+i)+j) = value; // if you deference (matrix + i) you get the row
// if you add the column and deference again, you
// get the actual item to store (not a pointer!)
}
One of the most important things to do now is to make sure we free the memory when we’re done. Each level of malloc() should have the same number of free() calls, and the calls should be in a FILO order (reverse of the malloc calls):
for(i=0; i<Rows; i++)
free(*(matrix + i));
free(matrix);
//set to NULL to clean up, matrix points to allocated memory now so let’s not use it!
matrix = NULL;
If you want to use a typedef'd array, it is even simpler.
Say you have in your code typedef int LabeledAdjMatrix[SIZE][SIZE];
You can then use:
LabeledAdjMatrix *pMatrix = malloc(sizeof(LabeledAdjMatrix));
Then you can write:
for (i=0; i<SIZE; i++) {
for (j=0; j<SIZE; j++) (*parr)[i][j] = k++; /* or parr[0][i][j]... */
}
Because pArr is a pointer to you matrix and * has lower priority than [];
That is why a mode common idiom is to typedef the row:
typedef int LabeledAdjRow[SIZE];
Then you can write:
LabeledAdjRow *pMatrix = malloc(sizeof(LabeledAdjRow) * SIZE);
for (i=0; i<SIZE; i++) {
for (j=0; j<SIZE; j++) parr[i][j] = k++;
}
And you can memcpy all that directly:
LabeledAdjRow *pOther = malloc(sizeof(LabeledAdjRow) * SIZE);
memcpy(pOther, pMatrix, sizeof(LabeledAdjRow) * SIZE);
Going off of Jen's answer, if you want to allocate space for a 3D array, in the same manner, you can do
int(*A)[n][n] = malloc(sizeof(int[num_of_2D_arrays][n][n]));
for (size_t p = 0; p < num_of_2D_arrays; p++)
for (size_t i = 0; i < n; i++)
for (size_t j = 0; j < n; j++)
A[p][i][j] = p;
for (size_t p = 0; p < num_of_2D_arrays; p++)
printf("Outter set %lu\n", p);
for (size_t i = 0; i < n; i++)
for (size_t j = 0; j < n; j++)
printf(" %d", A[p][i][j]);
printf("\n");
free(A);
my row size is fixed but the column size keeps varying all the time.That's why I am thinking to use pointers for column level.But I am not sure how to use this feature in C.Please kindly help me.
int array[rows][columns]; //row size fixed but column size is not fixed.
If all but one dimension of your array are fixed, which is the situation that you describe, you can avoid using arrays of pointers: you could typedef a row as int[rows], and create an array of rows, like this:
typedef int row_t[rows];
Now you can pass around 2D arrays composed of row_t arrays like this:
int total(row_t array[], int len) {
int res = 0;
for (int c = 0 ; c != len ; c++) {
for (int r = 0 ; r != rows ; r++) {
res += array[r][c];
}
}
return res;
}
int main() {
row_t *matrix = malloc(columns*sizeof(row_t));
for (int c = 0 ; c != columns ; c++) {
for (int r = 0 ; r != rows ; r++) {
matrix[r][c] = r*r+c*c;
}
}
printf("%d\n", total(matrix, columns));
free(matrix);
return 0;
}
You need to declare an array of pointers. Like this:
int *array[rows];
Later, according to the need, allocate memory to each row. You can use malloc or calloc functions for this.
for(int i=0; i<rows; i++){
array[i] = (int *) malloc(columns * sizeof(int));
}
This method also allows one to have rows with varying sizes.
you cannot use variable in array dimensions so use pointer to pointers
int **array;
array = (int **)malloc(sizeof(int *) * rows);
for (int i = 0; i < rows; ++i) {
array[i] = (int *)malloc(sizeof(int) * columns);
for (int j = 0; j < columns; j++) {
array[i][j] = value;
}
}
and ofcourse do not forget to free it all
You would do this by having an array of int*, e.g.
int* array[row];
for (int i = 0; i < row; ++i)
array[i] = (int*)malloc(i * sizeof(int));
And then you can access the data the same as if it was static.
int val = array[myRow][myColumn];
You'll have to be sure to free the memory in the same pattern as you allocated it.
Since 13 years now C allows dynamic sizes in the declaration of arrays. This is called variable length array, VLA.
If you really have large arrays, this might explose your stack. But you can still allocate such a beast with malloc and have all the simplicity of a 2D array:
size_t n = SOME_COMPLICATED_VALUE;
size_t m = ANOTHER_ONE;
double (*A)[n] = malloc(double[m][n]);
If each row can have a different number of columns, then the general approach is to declare an array of pointers, and then for each row dynamically allocate the number of elements for that row:
int *array[ROWS];
array[i] = malloc(sizeof *array[i] * number_of_columns_for_row_i);
Note that when you are done, you'll have to free each row individually:
for (i = 0; i < ROWS; i++)
free(array[i]);
If all the rows have the same number of columns, but that number can change for each instance of the array, then you have several choices. If you're using a C99-compliant compiler, you can use a variable length array, where the array dimension isn't known until runtime:
int columns;
// get number of columns somehow
int array[ROWS][columns];
VLAs are handy, but have their limitations; they cannot be members of struct or union types, and they cannot be declared static or at file scope. The big advantage of this approach is that you don't have to worry about deallocating anything when you're done. The big disadvantage of this approach is that you may not have enough memory to satsify the request; the memory space available for auto variables is typically quite limited compared to memory for dynamically-allocated objects. Also, the latest revision of the language standard (C11) makes VLAs optional, so they may not be available in all implementations going forward.
Alternately, you could dynamically allocate the array like so:
int (*array)[ROWS];
int columns;
// get number of columns somehow
array = malloc(sizeof *array * columns);
Note that this time, array is a pointer to an array of int, not an array of pointers to int. The benefit of this approach is that the array memory is allocated contiguously, and free-ing the memory is as simple as writing
free(array);
The drawback is that if your array is really big or the dynamic memory pool is really fragmented, you may not have a large enough chunk of memory available to satisfy the request.
If you don't need all the rows to be allocated contiguously, then use the first approach that allocates each row individually; just use the same value for the number of columns:
int *array[ROWS];
int columns;
// get number of columns
for (i = 0; i < ROWS; i++)
array[i] = malloc(sizeof *array[i] * columns);
I create a 2D array in C as follows:
int **arr;
arr = malloc(rows * sizeof(int *));
for (i = 0; i < rows; i++)
arr[i] = malloc(cols * sizeof(int));
Now, I call:
func(arr)
In the function func, how do I calculate the row and column dimensions?
You can't calculate it - arr is just a pointer to a pointer, there is no more information associated with it (as with all C arrays). You have to pass the dimensions as separate arguments.
You can't. You have to pass the dimensions along with your array to the function func(arr).
you can't (the beauty of C). (and don't try using sizeof, because that will only give you the size of the pointer) If another function needs to know the dimensions of the array, you'll have to pass along those parameters (height and width) as arguments along with the array pointer.
A workaround for this would be to not use an array directly, but instead have a struct like this:
struct table{
int ** arr;
int rows;
int columns;
}
You can then have a function which creates instances of table that takes the number of rows and columns, and handles the allocation.
As everyone else has said, you can't. However, you may find it useful to create a structure to contain the row, column and pointer all together. This will allow you to say:
typedef struct {
int rows;
int cols;
int **data;
} myDataType;
...
foo(myData);
void foo(myDataType myData) {
for( i = 0; i < myData.rows; i++) {
for( j = 0; j < myData.cols; j++ ) {
printf("%d,%d: %d\n", i, j, myData.data[i][j]);
}
}
}
(I apologize if my syntax is slightly off; Perl, Java, C# and a little Ruby are jockying for elbow-space.)
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];
}