How to allocate a 3 dimensional array in C? [duplicate] - c

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Malloc a 3-Dimensional array in C?
(16 answers)
Closed 8 years ago.
Creating two dimensional arrays in C is easy:
char (*arr)[50] = malloc(sizeof(arr) * 10 * 50); // 10x50 matrix
How do you do three dimensional arrays in C? It doesn't look like I can do something like:
char (**arr)[50] = malloc(sizeof(arr) * 10 * 20 * 50); // 10x20x50 matrix?

Answer 1

Three dimensional array requires 2 dimensions to be known
char (*arr)[20][50] = malloc(sizeof(char) * 10 * 20 * 50)
Note: I have corrected sizeof(arr) to sizeof(char), because sizeof(arr) will return the size of a pointer.

Answer 2

A possible way could be to allocate a mono-dimensional array, e.g.
int width=10; length=20; height=50;
char* arr = malloc(width*length*height);
if (!arr) { perror("malloc"); exit(EXIT_FAILURE); };
then have some way to access it, for instance a macro
#define Element(I,J,K) arr[width*length*(I)+length*(J)+(K)]
and use Element(i,j,k)
You could pack all this using a flexible array member like
struct my3dstring_st {
int width;
int length;
int height;
char arr[];
};
then have a.g. a making function
struct my3dstring_st *
make_my3dstring (int width, int length, int height)
{
if (width<=0 || length<=0 || height<=0) return NULL;
struct my3dstring_st* s =
malloc(sizeof(struct my3dstring_st)
+ width * length * height);
if (!s) {perror("malloc"); exit(EXIT_FAILURE); };
s->width = width;
s->length = length;
s->height = height;
memset (s->arr, 0, width * length * height);
return s;
}
and an inline accessing function (in a header file):
static inline int
access_m3dstring(struct my3dstring_st*s, int i, int j, int k) {
if (!s || i<0 || j<0 || k<0
|| i>=s->width || j>=s->height || k>=s->length) return EOF;
return s->arr[i*->width*s->height + j*s->height + k];
}
I leave as an exercise to write the modification function modify_m3dstring, and you could have unsafe but faster variants which don't do any checks...

Answer 3

General rules:
T *arr = malloc( sizeof *arr * n ); // for an N-element array
T (*arr)[N] = malloc( sizeof *arr * m ); // for an NxM-element array
T (*arr)[N][M] = malloc( sizeof *arr * k ); // for an NxMxK-element array
where uppercase letters indicate values that are known at compile time and lowercase letters indicate values that are known at run time. The pattern for higher-dimensional arrays should be obvious.
If you are using C99 compiler or a C2011 compiler that supports variable-length arrays, you can use runtime variables for all your dimensions:
size_t n = some_value();
size_t m = some_other_value();
size_t k = yet_another_value();
T (*arr)[n][m] = malloc( sizeof *arr * k );
The type of the expression *arr is T [n][m], so sizeof *arr gives the same result as sizeof (T) * n * m; the result is easier to read and less prone to errors.
If your compiler doesn't support VLAs and you don't know your dimensions at compile time, you'll either have to allocate as a 1-d array and compute offsets manually:
T *arr = malloc( sizeof *arr * n * m * k );
...
arr[ 3*n*m + 2*m + 1] = x; // equivalient to arr[3][2][1] = x
Or, if you can live with your rows not being adjacent in memory, you could allocate the array piecemeal:
T ***arr = malloc (sizeof *arr * n );
for (size_t i = 0; i < n; i++ )
{
arr[i] = malloc( sizeof *arr[i] * m );
for (size_t j = 0; j < m; j++ )
{
arr[i][j] = malloc( sizeof *arr[i][j] * k )
}
}
Ideally, you should check the result of each malloc to make sure it succeeded. And you'll have to free the array in the reverse order that you allocated it:

Answer 4

char (*arr)[20][50] = malloc(sizeof(char) * 10 * 20 * 50);
sizeof(char) is guaranteed to be 1. So, it can be omitted.
char (*arr)[20][50] = malloc(10 * 20 * 50);