// 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];
}
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.
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);
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.
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?
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 */
}
}
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.
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.
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.
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
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).
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));
}
Source: Stackoverflow.com