How to initialize a 3D contiguous array in C - c

I know how to do a potentioal non-contiguous array in the following way:
int main () {
int ***array = (int***)malloc(3*sizeof(int**));
int i, j;
for (i = 0; i < 3; i++) {
// Assign to array[i], not *array[i] (that would dereference an uninitialized pointer)
array[i] = (int**)malloc(3*sizeof(int*));
for (j = 0; j < 3; j++) {
array[i][j] = (int*)malloc(3*sizeof(int));
}
}
array[1][2][1] = 10;
return 0;
}
with the code above, the array[0][j] blocks can be not contiguous. To get contiguous, I feel that we need to malloc in this way
int* array = (int*)malloc(3*3*3*sizeof(int));
int** y = (int**)malloc(3*3*sizeof(int**));
int*** x = (int***)malloc(3*sizeof(int***));
for(i = 0; i < 3; i++)
{
vals = vals + i*m*n;
x[i] = &vals;
for(j = 0; j < 3; j++)
{
x[i][j] = vals + j * n;
}
}
However, I got troulbe with address assignment. I am not a c programmer, can anyone correct my fault? Thanks in advance...

int*** x = (int***)malloc(3*sizeof(int***));
should be
int*** x = (int***)malloc(3*sizeof(int**));
Now initialization can be :
for(i = 0; i < 3; i++)
{
x[i] = y + 3*i;
for(j = 0; j < 3; j++)
{
x[i][j] = array + i*3*3 + j*3;
}
}
So that x[0] points to the first element of y, x[1] to the fourth, etc.
And x[0][0]=y[0] to the first of array, x[0][1]=y[1] to the fourth of array, etc.

To allocate a contiguous 3D array, you only need to do the following (assumes all dimensions are known at compile time):
#define D0 ...
#define D1 ...
#define D2 ...
...
T (*arr)[D1][D2] = malloc( sizeof *arr * D0 ); // for any type T
...
arr[i][j][k] = some_value();
...
arr is declared as a pointer to a D1xD2 array. We then allocate enough space for D0 such arrays (sizeof *arr == sizeof (T [D1][D2])).
With this method, all of the memory for the array is allocated contiguously. Also, you only need one call to free to deallocate the whole thing.
If your dimensions are not known until runtime and you're using a C99 compiler or a C2011 compiler that supports variable-length arrays, you're still in luck:
size_t d0, d1, d2;
...
T (*arr)[d1][d2] = malloc( sizeof *arr * d0 );
The main issue is how to pass this as an argument to a function. Assuming that D1 and D2 are known at compile time, if you decide to pass it as
foo( arr, D0 );
then the prototype for foo will need to be
void foo( T (*aptr)[D1][D2], size_t n )
{
...
aptr[i][j][k] = ...;
}
and it will only be useful for n x D1 x D2-sized arrays.
If you go the VLA route, you'll need to declare the dimensions and pass values for them as well:
void foo( size_t d0, size_t d1, size_t d2, T (*aptr)[d1][d2] ) // d1 and d2 *must* be
// declared before aptr
{
...
arr[i][j][k] = some_value();
}
void bar( void )
{
size_t d0, d1, d2;
...
T (*arr)[d1][d2] = malloc( sizeof *arr * d0 );
...
foo( d0, d1, d2, arr );
...
}
If you don't have a compiler that supports VLAs, but you still want to allocate this memory contiguously, then you'll have to go the old-fashioned route - allocate it as a 1D array and compute your offsets manually:
T *arr = malloc( sizeof *arr * d0 * d1 * d2 );
...
arr[i * d0 * d1 + j * d1 + k] = some_value();

I am using some pretty neat methods for allocating multi-dimensional arrays with row pointers. The functions multi_malloc and multi_free can be used for arrays with arbitrary dimensions and arbitrary types and they work on basically all platforms and from C and C++
You can allocate, e.g. a 3-dimensional array with row-pointers recursively. E.g. a 10x20x30 dimensional array
float*** data = (float***) multi_malloc(sizeof(float),3, 10,20,20);
access elements like
data[2][3][4] = 2.0;
and free everything like (data as well as row pointers)
multi_free(data,3);
The header, which I think should be part of C is
#pragma once
#include <stdlib.h>
#include <stddef.h>
#include <stdarg.h>
#include <string.h>
#if (defined(_MSC_VER) && defined(_WIN32))
// Note when used inside a namespace, the static is superfluous
# define STATIC_INLINE_BEGIN static inline //__forceinline
# define STATIC_INLINE_END
#elif (defined(__GNUC__))
# define STATIC_INLINE_BEGIN static inline
# if defined(__CYGWIN__)
# define STATIC_INLINE_END
# else
# define STATIC_INLINE_END __attribute__ ((always_inline))
# endif
#endif
STATIC_INLINE_BEGIN void* multi_malloc(size_t s, size_t d, ...) STATIC_INLINE_END;
STATIC_INLINE_BEGIN void multi_free(void *r, size_t d) STATIC_INLINE_END;
/**
* Allocate multi-dimensional array and establish row pointers
*
* #param s size of each element
* #param d number of dimension
*
* #return
*/
STATIC_INLINE_BEGIN void* multi_malloc(size_t s, size_t d, ...) {
char* tree;
va_list ap; /* varargs list traverser */
size_t max, /* size of array to be declared */
*q; /* pointer to dimension list */
char **r, /* pointer to beginning of the array of the
* pointers for a dimension */
**s1, *t; /* base pointer to beginning of first array */
size_t i, j; /* loop counters */
size_t *d1; /* dimension list */
va_start(ap,d);
d1 = (size_t *) malloc(d*sizeof(size_t));
for(i=0;i<d;i++)
d1[i] = va_arg(ap,size_t);
r = &tree;
q = d1; /* first dimension */
if (d==1) {
max = *q;
free(d1);
return malloc(max*d);
}
max = 1;
for (i = 0; i < d - 1; i++, q++) { /* for each of the dimensions
* but the last */
max *= (*q);
r[0]=(char *)malloc(max * sizeof(char **));
r = (char **) r[0]; /* step through to beginning of next
* dimension array */
}
max *= s * (size_t) (*q); /* grab actual array memory */
r[0] = (char *)malloc(max * sizeof(char));
/*
* r is now set to point to the beginning of each array so that we can
* use it to scan down each array rather than having to go across and
* then down
*/
r = (char **) tree; /* back to the beginning of list of arrays */
q = d1; /* back to the first dimension */
max = 1;
for (i = 0; i < d - 2; i++, q++) { /* we deal with the last
* array of pointers later on */
max *= (*q); /* number of elements in this dimension */
for (j=1, s1=r+1, t=r[0]; j<max; j++) { /* scans down array for
* first and subsequent
* elements */
/* modify each of the pointers so that it points to
* the correct position (sub-array) of the next
* dimension array. s1 is the current position in the
* current array. t is the current position in the
* next array. t is incremented before s1 is, but it
* starts off one behind. *(q+1) is the dimension of
* the next array. */
*s1 = (t += sizeof (char **) * *(q + 1));
s1++;
}
r = (char **) r[0]; /* step through to begining of next
* dimension array */
}
max *= (*q); /* max is total number of elements in the
* last pointer array */
/* same as previous loop, but different size factor */
for (j = 1, s1 = r + 1, t = r[0]; j < max; j++)
*s1++ = (t += s * *(q + 1));
va_end(ap);
free(d1);
return((void *)tree); /* return base pointer */
}
/**
* Free multi-dimensional array and corresponding row pointers
*
* #param r data
* #param d number of dimensions
*/
STATIC_INLINE_BEGIN void multi_free(void *r, size_t d) {
void **p;
void *next=NULL;
size_t i;
for (p = (void **)r, i = 0; i < d; p = (void **) next,i++)
if (p != NULL) {
next = *p;
free(p);
p = NULL;
}
}

You can allocate memory for buffer of items where each item is a two dimensional array. So it is effectively is a three dimensional array:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 3
int main()
{
int (*array)[N][N] = malloc(N * N * N * sizeof(int));
/* set 0 to all values */
memset(array, 0, N * N * N * sizeof(int));
/* use as 3D array */
array[0][0][0] = 1;
array[1][1][1] = 2;
array[2][2][2] = 3;
int i;
/* print array as contiguous buffer */
for (i = 0; i < N * N * N; ++i)
printf("i: %d\n", ((int*) array)[i]);
free(array);
return 0;
}
So, in memory the array is placed as regular int array[N][N][N].

Although I think a normal array, created on the stack would be best:
int array[3][3][3];//can avoid a lot of free() calls later on
Here is a way to create a 3D array dynamically:
(I use calloc here instead of malloc as it creates initialized memory space)
int *** Create3D(int p, int c, int r)
{
int ***arr;
int x,y;
arr = calloc(p, sizeof(arr)); //memory for int
for(x = 0; x < p; x++)
{
arr[x] = calloc(c ,sizeof(arr)); //memory for pointers
for(y = 0; y < c; y++)
{
arr[x][y] = calloc(r, sizeof(int));
}
}
return arr;
}
Usage could be:
int ***array = Create3D(3,3,3);
for(i=0;i<3;i++)
for(j=0;j<3;j++)
for(k=0;k<3;k++)
array[i][j][k] = (i+1)*(j+1)*(k+1);
Note that the return of [c][m][re]alloc() is not cast in this example. Although not strictly forbidden in C, it is not recommended. (this is not the case in C++, where it is required)
Keep in mind, everything allocated, must be freed. Notice freeing is done in reverse order of allocating:
void free3D(int ***arr, int p, int c)
{
int i,j;
for(i=0;i<p;i++)
{
for(j=0;j<c;j++)
{
if(arr[i][j]) free(arr[i][j]);
}
if(arr[i]) free(arr[i]);
}
if(arr) free(arr);
}
Usage could be:
free3D(array,3,3);

Related

What exactly is this method of contiguous C memory allocation doing under the hood?

I came across this question while looking for an effective way to allocate large multi-dimensional arrays contiguously in memory. The accepted answer suggests that for a 3D array of size sz[0] x sz[1] x sz[2] one should use this method, which is currently melting my feeble brain:
int (*a)[sz[1]][sz[2]] = calloc(sz[0], sizeof(*a));
...
free(a)
the left hand of that statement looks like a 2D array of int * allocated on the stack. The right side is a single (?!) call to calloc() that allocates int * on the heap. Since sizeof(*a)==sizeof(int *) (right?) this looks like too few allocations to make any sense, since it appears to allocate sz[0]x int * bytes, and yet it works to index over the full intended size of the array.
Can someone please help me understand how exactly this definition works to produce the intended result? Is the C compiler repeating the call to calloc for every entry in the table defined on the left? And if so, how does a single call to free() suffice to get rid of it? Does the resulting array reside entirely on the heap, or is it mixing a reference table on the stack that points to memory allocated on the heap?
Here is some code with a similar principle that maybe is easier to understand at first:
typedef int THING[5][6]; // THING means a contiguous array of 5x6 ints
THING arr[4]; // arr is a contiguous array of 4 THINGs
THING *first = &arr[0]; // The expression *first would yield the first thing.
Hopefully you recognize the last two lines here as being common syntax for non-dynamic allocation of any array, and referring to the array's first element. That works just the same whether or not THING is itself an array.
Now, &arr[0] points to a memory location that is the start of a contiguous block of ints of size 4x5x6. if you use dynamic allocation to make that block it looks like:
THING *first = malloc( sizeof(int[4][5][6]) );
If we expand out the typedef in this last line it looks like:
int (*first)[5][6] = malloc( sizeof(int[4][5][6]) );
The code in your question is the same as this last line , except that:
it uses variables instead of hardcoded integers (which is allowed since C99).
it uses calloc instead of malloc.
it uses a more robust syntax for calculating the size to allocate, see here for explanation.
To not rely on VLAs, but still use one continuous region of memory you could use this approach:
int *** int_array_3d_allocate(size_t x, size_t y, size_t z)
{
int *** result;
size_t n = x;
size_t s = n * sizeof *result; /* x vector of pointer to int** */
n *= y;
s += n * sizeof **result; /* x*y vectors of pointer to int* */
n *= z;
s += n * sizeof ***result; /* x*y*z int */
/* allocate it */
result = malloc(s);
if (result)
{
/* make the int** vector point to the int* vectors: */
for (size_t i = 0; i < x; ++i)
{
result[i] = (int**) ((char*) result) +
(x * sizeof *result +
i * y * sizeof **result);
}
/* make the int* vectors point to the int vectors: */
for (size_t i = 0; i < x*y; ++i)
{
((int**) ((char*) result + x * sizeof *result))[i] = (int*) ((char*) result) +
(x * sizeof *result + x*y * sizeof **result
+ i * sizeof ***result);
}
}
return result;
}
Version of the above code taking care of correct alignment of the int* and the int** blocks:
#include <stdalign.h>
int *** int_array_3d_allocate(size_t x, size_t y, size_t z)
{
int *** result;
size_t n = x;
size_t s = n * sizeof *result; /* x vector of pointer to int** */
size_t y_off = s % alignof **result
?alignof **result - s % alignof **result :0;
n *= y;
s += n * sizeof **result; /* x*y vectors of pointer to int* */
size_t z_off = s % alignof ***result
?alignof ***result - s % alignof ***result :0;
n *= z;
s += n * sizeof ***result; /* x*y*z int */
/* allocate it */
result = malloc(s);
if (result)
{
/* make the int** vector point to the int* vectors: */
for (size_t i = 0; i < x; ++i)
{
result[i] = (int**) ((char*) result) + y_off +
(x * sizeof *result +
i * y * sizeof **result);
}
/* make the int* vectors point to the int vectors: */
for (size_t i = 0; i < x*y; ++i)
{
((int**) ((char*) result + x * sizeof *result + y_off))[i] = (int*) ((char*) result) + y_off +
(x * sizeof *result + x*y * sizeof **result + z_off +
+ i * sizeof ***result);
}
}
return result;
}
Use it like this:
#include <stdlib.h>
#include <stdio.h>
int *** int_array_3d_allocate(size_t x, size_t y, size_t z);
int main(void)
{
const size_t x = 2;
const size_t y = 3;
const size_t z = 5;
int *** int_array_3d = int_array_3d_allocate(x, y, z);
if (!int_array_3d)
{
perror("int_array_3d_allocate() failed");
}
else
{
for (size_t i = 0; i < x; ++i)
{
for (size_t j = 0; j < y; ++j)
{
for (size_t k = 0; k < z; ++k)
{
int_array_3d[i][j][k] = (int)(i*j*k);
}
}
}
/* do stuff with the continuous array of ints.
Just be aware that the 1st int only is located at address:
(char* int_array_3d) +
(x * sizeof *int_array_3d + x*y * sizeof **int_array_3d)
*/
free(int_array_3d);
}
}
If the array is handed off to functions, it decays into a pointer-to-pointer-to-pointer-to-int, making it unwieldy; one has to pass all the extra size information, too, or pass a pointer to a fixed size; see What is array decaying? A different way of handling arrays with multiple dimensions is an object which has the dimensions encoded within the object. This will compile in C90,
#include <stdlib.h> /* mallc, free, EXIT_ */
#include <errno.h> /* errno */
#include <stdio.h> /* perror, printf, fput[c|s] */
struct IntCube { size_t x, y, z; /* C99 supports FAM; would be useful. */ };
/** Returns a `struct IntCube` with `x`, `y`, `z` dimensions or null and
`errno` may be set. The caller is responsible for calling `free`. */
static struct IntCube *IntCube(const size_t x, const size_t y, const size_t z) {
struct IntCube *cube;
size_t xy_size, xyz_size, data_size, cube_size;
if(!x || !y || !z) return 0;
/* Check for overflow; <https://stackoverflow.com/q/1815367/2472827>. */
xy_size = x * y;
xyz_size = xy_size * z;
data_size = xyz_size * sizeof(int);
cube_size = sizeof cube + data_size;
if(xy_size / x != y
|| xyz_size / xy_size != z
|| data_size / xyz_size != sizeof(int)
|| cube_size < data_size) { errno = ERANGE; return 0; }
/* Allocate memory. */
if(!(cube = malloc(cube_size))) return 0; /* POSIX has defined errors. */
cube->x = x;
cube->y = y;
cube->z = z;
return cube;
}
static int *int_cube_get(const struct IntCube *cube,
const size_t x, const size_t y, const size_t z) {
return (int *)(cube + 1) + z * cube->y * cube->x + y * cube->x + x;
}
typedef void (*IntCubeAction)(const size_t x, const size_t y, const size_t z,
int *pnumber);
typedef void (*BinaryAction)(int bin);
/** Goes through `cube` and performs `action` on each number. It will call
optional binary action `bin` each time there is an
start(false)/end(true)-of-x/y. */
static void IntCubeForEach(struct IntCube *const cube,
const IntCubeAction action, const BinaryAction bin) {
size_t x, y, z;
if(!cube || !action) return;
for(z = 0; z < cube->z; z++) {
if(bin) bin(0);
for(y = 0; y < cube->y; y++) {
if(bin) bin(0);
for(x = 0; x < cube->x; x++) {
action(x, y, z, int_cube_get(cube, x, y, z));
}
if(bin) bin(1);
}
if(bin) bin(1);
}
}
/** #implements IntCubeAction */
static void fill_with_xyz(const size_t x, const size_t y, const size_t z,
int *pnumber) {
*pnumber = (x + 1) * (y + 1) * (z + 1);
}
/** #implements IntCubeAction */
static void print_cube(const size_t x, const size_t y, const size_t z,
int *pnumber) {
(void)y, (void)z;
printf("%s%d", x ? ", " : "", *pnumber);
}
/** #implements BinaryAction */
static void print_cube_corners(int bin) {
printf("%s", bin ? " }" : "{ ");
}
int main(void) {
struct IntCube *cube = 0;
int status = EXIT_FAILURE;
if(!(cube = IntCube(4, 3, 3))) goto catch;
IntCubeForEach(cube, &fill_with_xyz, 0);
IntCubeForEach(cube, &print_cube, &print_cube_corners);
fputc('\n', stdout);
status = EXIT_SUCCESS;
goto finally;
catch:
perror("Cube");
finally:
free(cube);
return status;
}
{ { 1, 2, 3, 4 }{ 2, 4, 6, 8 }{ 3, 6, 9, 12 } }{ { 2, 4, 6, 8 }{ 4, 8, 12, 16 }{ 6, 12, 18, 24 } }{ { 3, 6, 9, 12 }{ 6, 12, 18, 24 }{ 9, 18, 27, 36 } }
This creates a dependence on struct IntCube, but with the dependence, one can calculate the size at runtime.

sort array in C, return sorted indices

I'm using an example from https://phoxis.org/2012/07/12/get-sorted-index-orderting-of-an-array/ where he returns the sort indices from a sort of an array, i.e.
3,4,2,6,8 returns 4,3,1,0,2 (+1 for each index in R). This is the equivalent of R's order function
I've translated his/her code to work as a function returning an array of sorted indices. The code gives the correct answer.
keeping track of the original indices of an array after sorting in C has a similar response, but as #BLUEPIXY warns, his solution doesn't work in all circumstances. I need something that will work in all circumstances, including ties.
however, the original author uses a global pointer, which causes a memory leak, and free() doesn't fix it. which I don't know how to do this without the global pointer.
How can I fix this memory leak, or at least return sorted indices in C that will always work?
#include <stdio.h>
#include <stdlib.h>
/* holds the address of the array of which the sorted index
* order needs to be found
*/
int * base_arr = NULL;
/* Note how the compare function compares the values of the
* array to be sorted. The passed value to this function
* by `qsort' are actually the `idx' array elements.
*/
static int compar_increase (const void * a, const void * b) {
int aa = *((int * ) a), bb = *((int *) b);
if (base_arr[aa] < base_arr[bb]) {
return 1;
} else if (base_arr[aa] == base_arr[bb]) {
return 0;
} else {
// if (base_arr[aa] > base_arr[bb])
return -1;
}
}
int * order_int (const int * ARRAY, const size_t SIZE) {
int * idx = malloc(SIZE * sizeof(int));
base_arr = malloc(sizeof(int) * SIZE);
for (size_t i = 0; i < SIZE; i++) {
base_arr[i] = ARRAY[i];
idx[i] = i;
}
qsort(idx, SIZE, sizeof(int), compar_increase);
free(base_arr); base_arr = NULL;
return idx;
}
int main () {
const int a[] = {3,4,2,6,8};
int * b = malloc(sizeof(int) * sizeof(a) / sizeof (*a));
b = order_int(a, sizeof(a) / sizeof(*a));
for (size_t i = 0; i < sizeof(a)/sizeof(*a); i++) {
printf("b[%lu] = %d\n", i, b[i]+1);
}
free(b); b = NULL;
return 0;
}
A straightforward approach without using a global variable can look the following way
#include <stdio.h>
#include <stdlib.h>
int cmp_ptr(const void *a, const void *b)
{
const int **left = (const int **)a;
const int **right = (const int **)b;
return (**left < **right) - (**right < **left);
}
size_t * order_int(const int *a, size_t n)
{
const int **pointers = malloc(n * sizeof(const int *));
for (size_t i = 0; i < n; i++) pointers[i] = a + i;
qsort(pointers, n, sizeof(const int *), cmp_ptr);
size_t *indices = malloc(n * sizeof(size_t));
for (size_t i = 0; i < n; i++) indices[i] = pointers[i] - a;
free(pointers);
return indices;
}
int main( void )
{
const int a[] = { 3,4,2,6,8 };
const size_t N = sizeof(a) / sizeof(*a);
size_t *indices = order_int(a, N);
for (size_t i = 0; i < N; i++) printf("%d ", a[indices[i]]);
putchar('\n');
free(indices);
return 0;
}
The program output is
8 6 4 3 2
As for the memory leak then it is due to overwriting the value of the pointer to redundantly allocated memory.
int * b = malloc(sizeof(int) * sizeof(a) / sizeof (*a));
b = order_int(a, sizeof(a) / sizeof(*a));
The memory allocation does not make sense.
The problem I see is that within main function - you are allocating pointer b some memory -
int * b = malloc(sizeof(int) * sizeof(a) / sizeof (*a));
The next line calls order_int(...) that returns a pointer to already allocated memory -
b = order_int(a, sizeof(a) / sizeof(*a));
Looking at the order_int function -
int * order_int (const int * ARRAY, const size_t SIZE) {
int * idx = malloc(SIZE * sizeof(int));
base_arr = malloc(sizeof(int) * SIZE);
for (size_t i = 0; i < SIZE; i++) {
base_arr[i] = ARRAY[i];
idx[i] = i;
}
qsort(idx, SIZE, sizeof(int), compar_increase);
free(base_arr); base_arr = NULL;
return idx;
}
.. you see that idx has been already been allocated the correct memory.
I would suggest removing the malloc from b - see below.
int * b = NULL;

Assign Memory to 3D array using triple pointer

I have to assign memory to a 3D array using a triple pointer.
#include <stdio.h>
int main()
{
int m=10,n=20,p=30;
char ***z;
z = (char***) malloc(sizeof(char**)*m*n*p);
return 0;
}
Is this correct way of doing this?(I think what i am doing is incorrect.)
To completely allocate a 3D dynamic array you need to do something like the following:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int m=10,n=20,p=30;
char ***z;
z = malloc(m * sizeof(char **));
assert(z != NULL);
for (i = 0; i < m; ++i)
{
z[i] = malloc(n * sizeof(char *));
assert(z[i] != NULL);
for (j = 0; j < n; ++j)
{
z[i][j] = malloc(p);
assert(z[i][j] != NULL);
}
}
return 0;
}
Freeing the data is left as an exercise for the reader.
There's no need to cast the return value of malloc(), in C.
And if you expect to store m * n * p characters directly (and compute the address yourself), then you should of course not scale the allocation by the size of a char **.
You mean:
int m = 10, n = 20, p = 30;
char *z = malloc(m * n * p * sizeof *z);
This will allocate 10 * 20 * 30 = 6000 bytes. This can be viewed as forming a cube of height p, with each "slice" along the vertical axis being n * m bytes.
Since this is for manual addressing, you cannot use e.g. z[k][j][i] to index, instead you must use z[k * n * m + j * m + i].
If you don't need the memory to be allocated in a single, contiguous chunk (which IME is the usual case), you would do something like this:
char ***z;
z = malloc(sizeof *z * m); // allocate m elements of char **
if (z)
{
int i;
for (i = 0; i < m; i++)
{
z[i] = malloc(sizeof *z[i] * n); // for each z[i],
if (z[i]) // allocate n elements char *
{
int j;
for (j = 0; j < n;j++)
{
z[i][j] = malloc(sizeof *z[i][j] * p); // for each z[i][j],
if (z[i][j]) // allocate p elements of char
{
// initialize each of z[i][j][k]
}
}
}
}
}
Note that you will need to free this memory in reverse order:
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
free(z[i][j];
free(z[i]);
}
free(z);
If you really need the memory to be allocated in a contiguous chunk, you have a couple of choices. You could allocate a single block and compute your offsets manually:
char *z = malloc(sizeof *z * m * n * p); // note type of z!
...
z[i * m + j * n + k] = some_value();
When you're done, you just need to do a single free:
free(z);
If you have a C99 compiler or a C11 compiler that supports variable-length arrays, you could do something like this:
int m=..., n=..., p=...;
char (*z)[n][p] = malloc(sizeof *z * m);
This declares z as a pointer to an nxp array of char, and we allocate m such elements. The memory is allocated contiguously and you can use normal 3-d array indexing syntax (z[i][j][k]). Like the above method, you only need a single free call:
free(z);
If you don't have a C99 compiler or a C11 compiler that supports VLAs, you would need to make n, and p compile-time constants, such as
#define n 20
#define p 30
otherwise that last method won't work.
Edit
m doesn't need to be a compile-time constant in this case, just n and p.
You would need the following nested loop -
z = (char**)malloc(sizeof(char*) * m);
for (int i = 0; i < m; ++i)
{
*(z + i) = (char*)malloc(sizeof(char*) * n);
for (int j = 0; j < n; ++j)
{
*(*(z + i)) = (char)malloc(p);
}
}
May not be synactically accurate, but it should be something along these lines.
You want sizeof(char) not sizeof(char**) as the latter will give you the size of a pointer which on most modern systems will be 4 bytes instead of the 1 you're expecting.

How do we allocate a 2-D array using One malloc statement

I have been asked in an interview how do i allocate a 2-D array and below was my solution to it.
#include <stdlib.h>
int **array;
array = malloc(nrows * sizeof(int *));
for(i = 0; i < nrows; i++)
{
array[i] = malloc(ncolumns * sizeof(int));
if(array[i] == NULL)
{
fprintf(stderr, "out of memory\n");
exit or return
}
}
I thought I had done a good job but then he asked me to do it using one malloc() statement not two. I don't have any idea how to achieve it.
Can anyone suggest me some idea to do it in single malloc()?
Just compute the total amount of memory needed for both nrows row-pointers, and the actual data, add it all up, and do a single call:
int **array = malloc(nrows * sizeof *array + (nrows * (ncolumns * sizeof **array));
If you think this looks too complex, you can split it up and make it a bit self-documenting by naming the different terms of the size expression:
int **array; /* Declare this first so we can use it with sizeof. */
const size_t row_pointers_bytes = nrows * sizeof *array;
const size_t row_elements_bytes = ncolumns * sizeof **array;
array = malloc(row_pointers_bytes + nrows * row_elements_bytes);
You then need to go through and initialize the row pointers so that each row's pointer points at the first element for that particular row:
size_t i;
int * const data = array + nrows;
for(i = 0; i < nrows; i++)
array[i] = data + i * ncolumns;
Note that the resulting structure is subtly different from what you get if you do e.g. int array[nrows][ncolumns], because we have explicit row pointers, meaning that for an array allocated like this, there's no real requirement that all rows have the same number of columns.
It also means that an access like array[2][3] does something distinct from a similar-looking access into an actual 2d array. In this case, the innermost access happens first, and array[2] reads out a pointer from the 3rd element in array. That pointer is then treatet as the base of a (column) array, into which we index to get the fourth element.
In contrast, for something like
int array2[4][3];
which is a "packed" proper 2d array taking up just 12 integers' worth of space, an access like array[3][2] simply breaks down to adding an offset to the base address to get at the element.
int **array = malloc (nrows * sizeof(int *) + (nrows * (ncolumns * sizeof(int)));
This works because in C, arrays are just all the elements one after another as a bunch of bytes. There is no metadata or anything. malloc() does not know whether it is allocating for use as chars, ints or lines in an array.
Then, you have to initialize:
int *offs = &array[nrows]; /* same as int *offs = array + nrows; */
for (i = 0; i < nrows; i++, offs += ncolumns) {
array[i] = offs;
}
Here's another approach.
If you know the number of columns at compile time, you can do something like this:
#define COLS ... // integer value > 0
...
size_t rows;
int (*arr)[COLS];
... // get number of rows
arr = malloc(sizeof *arr * rows);
if (arr)
{
size_t i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < COLS; j++)
arr[i][j] = ...;
}
If you're working in C99, you can use a pointer to a VLA:
size_t rows, cols;
... // get rows and cols
int (*arr)[cols] = malloc(sizeof *arr * rows);
if (arr)
{
size_t i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < cols; j++)
arr[i][j] = ...;
}
How do we allocate a 2-D array using One malloc statement (?)
No answers, so far, allocate memory for a true 2D array.
int **array is a pointer to pointer to int. array is not a pointer to a 2D array.
int a[2][3] is an example of a true 2D array or array 2 of array 3 of int
To allocate memory for a true 2D array, with C99, use malloc() and save to a pointer to a variable-length array (VLA)
// Simply allocate and initialize in one line of code
int (*c)[nrows][ncolumns] = malloc(sizeof *c);
if (c == NULL) {
fprintf(stderr, "out of memory\n");
return;
}
// Use c
(*c)[1][2] = rand();
...
free(c);
Without VLA support, if the dimensions are constants, code can use
#define NROW 4
#define NCOL 5
int (*d)[NROW][NCOL] = malloc(sizeof *d);
You should be able to do this with (bit ugly with all the casting though):
int** array;
size_t pitch, ptrs, i;
char* base;
pitch = rows * sizeof(int);
ptrs = sizeof(int*) * rows;
array = (int**)malloc((columns * pitch) + ptrs);
base = (char*)array + ptrs;
for(i = 0; i < rows; i++)
{
array[i] = (int*)(base + (pitch * i));
}
I'm not a fan of this "array of pointers to array" to solve the multi dimension array paradigm. Always favored a single dimension array, at access the element with array[ row * cols + col]? No problems encapsulating everything in a class, and implementing a 'at' method.
If you insist on accessing the members of the array with this notation: Matrix[i][j], you can do a little C++ magic. #John solution tries to do it this way, but he requires the number of column to be known at compile time. With some C++ and overriding the operator[], you can get this completely:
class Row
{
private:
int* _p;
public:
Row( int* p ) { _p = p; }
int& operator[](int col) { return _p[col]; }
};
class Matrix
{
private:
int* _p;
int _cols;
public:
Matrix( int rows, int cols ) { _cols=cols; _p = (int*)malloc(rows*cols ); }
Row operator[](int row) { return _p + row*_cols; }
};
So now, you can use the Matrix object, for example to create a multiplication table:
Matrix mtrx(rows, cols);
for( i=0; i<rows; ++i ) {
for( j=0; j<rows; ++j ) {
mtrx[i][j] = i*j;
}
}
You should now that the optimizer is doing the right thing and there is no call function or any other kind of overhead. No constructor is called. As long as you don't move the Matrix between function, even the _cols variable isn't created. The statement mtrx[i][j] basically does mtrx[i*cols+j].
It can be done as follows:
#define NUM_ROWS 10
#define NUM_COLS 10
int main(int argc, char **argv)
{
char (*p)[NUM_COLS] = NULL;
p = malloc(NUM_ROWS * NUM_COLS);
memset(p, 81, NUM_ROWS * NUM_COLS);
p[2][3] = 'a';
for (int i = 0; i < NUM_ROWS; i++) {
for (int j = 0; j < NUM_COLS; j++) {
printf("%c\t", p[i][j]);
}
printf("\n");
}
} // end of main
You can allocate (row*column) * sizeof(int) bytes of memory using malloc.
Here is a code snippet to demonstrate.
int row = 3, col = 4;
int *arr = (int *)malloc(row * col * sizeof(int));
int i, j, count = 0;
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
*(arr + i*col + j) = ++count; //row major memory layout
for (i = 0; i < r; i++)
for (j = 0; j < c; j++)
printf("%d ", *(arr + i*col + j));

dynamic allocating array of arrays in C

I don't truly understand some basic things in C like dynamically allocating array of arrays.
I know you can do:
int **m;
in order to declare a 2 dimensional array (which subsequently would be allocated using some *alloc function). Also it can be "easily" accessed by doing *(*(m + line) + column). But how should I assign a value to an element from that array? Using gcc the following statement m[line][column] = 12; fails with a segmentation fault.
Any article/docs will be appreciated. :-)
The m[line][column] = 12 syntax is ok (provided line and column are in range).
However, you didn't write the code you use to allocate it, so it's hard to get whether it is wrong or right. It should be something along the lines of
m = (int**)malloc(nlines * sizeof(int*));
for(i = 0; i < nlines; i++)
m[i] = (int*)malloc(ncolumns * sizeof(int));
Some side-notes:
This way, you can allocate each line with a different length (eg. a triangular array)
You can realloc() or free() an individual line later while using the array
You must free() every line, when you free() the entire array
Your syntax m[line][colummn] is correct. But in order to use a 2D array in C, you must allocate memory for it. For instance this code will allocated memory for a table of given line and column.
int** AllocateArray(int line, int column) {
int** pArray = (int**)malloc(line*sizeof(int*));
for ( int i = 0; i < line; i++ ) {
pArray[i] = (int*)malloc(column*sizeof(int));
}
return pArray;
}
Note, I left out the error checks for malloc for brevity. A real solution should include them.
It's not a 2d array - it's an array of arrays - thus it needs the multiple allocations.
Here's a modified version of quinmars' solution which only allocates a single block of memory and can be used with generic values by courtesy of void *:
#include <stdlib.h>
#include <string.h>
#include <assert.h>
void ** array2d(size_t rows, size_t cols, size_t value_size)
{
size_t index_size = sizeof(void *) * rows;
size_t store_size = value_size * rows * cols;
char * a = malloc(index_size + store_size);
if(!a) return NULL;
memset(a + index_size, 0, store_size);
for(size_t i = 0; i < rows; ++i)
((void **)a)[i] = a + index_size + i * cols * value_size;
return (void **)a;
}
int printf(const char *, ...);
int main()
{
int ** a = (int **)array2d(5, 5, sizeof(int));
assert(a);
a[4][3] = 42;
printf("%i\n", a[4][3]);
free(a);
return 0;
}
I'm not sure if it's really safe to cast void ** to int ** (I think the standard allows for conversions to take place when converting to/from void * ?), but it works in gcc. To be on the safe side, you should replace every occurence of void * with int * ...
The following macros implement a type-safe version of the previous algorithm:
#define alloc_array2d(TYPE, ROWS, COLS) \
calloc(sizeof(TYPE *) * ROWS + sizeof(TYPE) * ROWS * COLS, 1)
#define init_array2d(ARRAY, TYPE, ROWS, COLS) \
do { for(int i = 0; i < ROWS; ++i) \
ARRAY[i] = (TYPE *)(((char *)ARRAY) + sizeof(TYPE *) * ROWS + \
i * COLS * sizeof(TYPE)); } while(0)
Use them like this:
int ** a = alloc_array2d(int, 5, 5);
init_array2d(a, int, 5, 5);
a[4][3] = 42;
Although I agree with the other answers, it is in most cases better to allocate the whole array at once, because malloc is pretty slow.
int **
array_new(size_t rows, size_t cols)
{
int **array2d, **end, **cur;
int *array;
cur = array2d = malloc(rows * sizeof(int *));
if (!array2d)
return NULL;
array = malloc(rows * cols * sizeof(int));
if (!array)
{
free(array2d);
return NULL;
}
end = array2d + rows;
while (cur != end)
{
*cur = array;
array += cols;
cur++;
}
return array2d;
}
To free the array simply do:
free(*array); free(array);
Note: this solution only works if you don't want to change the order of the rows, because you could then lose the address of the first element, which you need to free the array later.
Humm. How about old fashion smoke and mirrors as an option?
#define ROWS 5
#define COLS 13
#define X(R, C) *(p + ((R) * ROWS) + (C))
int main(void)
{
int *p = (int *) malloc (ROWS * COLS * sizeof(int));
if (p != NULL)
{
size_t r;
size_t c;
for (r = 0; r < ROWS; r++)
{
for (c = 0; c < COLS; c++)
{
X(r,c) = r * c; /* put some silly value in that position */
}
}
/* Then show the contents of the array */
for (r = 0; r < ROWS; r++)
{
printf("%d ", r); /* Show the row number */
for (c = 0; c < COLS; c++)
{
printf("%d", X(r,c));
}
printf("\n");
}
free(p);
}
else
{
/* issue some silly error message */
}
return 0;
}
Using malloc(3) for allocate the first array and putting in there pointers created by malloc(3) should work with array[r][c] because it should be equivalent to *(*(array + r) + c), it is in the C standard.

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