malloc in C, but use multi-dimensional array synta

问题:

Is there any way to malloc a large array, but refer to it with 2D syntax? I want something like:

int *memory = (int *)malloc(sizeof(int)*400*200);
int MAGICVAR = ...;
MAGICVAR[20][10] = 3; //sets the (200*20 + 10)th element

---

UPDATE: This was important to mention: I just want to have one contiguous block of memory. I just don't want to write a macro like:

#define INDX(a,b) (a*200+b);

and then refer to my blob like:

memory[INDX(a,b)];

I'd much prefer:

memory[a][b];

---

UPDATE: I understand the compiler has no way of knowing as-is. I'd be willing to supply extra information, something like:

int *MAGICVAR[][200] = memory;

Does no syntax like this exist? Note the reason I don't just use a fixed width array is that it is too big to place on the stack.

---

UPDATE: OK guys, I can do this:

void toldyou(char MAGICVAR[][286][5]) {
  //use MAGICVAR
}

//from another function:
  char *memory = (char *)malloc(sizeof(char)*1820*286*5);
  fool(memory);

I get a warning, passing arg 1 of toldyou from incompatible pointer type, but the code works, and I've verified that the same locations are accessed. Is there any way to do this without using another function?

回答1:

Yes, you can do this, and no, you don't need another array of pointers like most of the other answers are telling you. The invocation you want is just:

int (*MAGICVAR)[200] = malloc(400 * sizeof *MAGICVAR);
MAGICVAR[20][10] = 3; // sets the (200*20 + 10)th element

---

If you wish to declare a function returning such a pointer, you can either do it like this:

int (*func(void))[200]
{
    int (*MAGICVAR)[200] = malloc(400 * sizeof *MAGICVAR);
    MAGICVAR[20][10] = 3;

    return MAGICVAR;
}

Or use a typedef, which makes it a bit clearer:

typedef int (*arrayptr)[200];

arrayptr function(void)
{
    /* ... */

回答2:

Use a pointer to arrays:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int (*arr)[10];

    arr = malloc(10*10*sizeof(int));
    for (int i = 0; i < 10; i++)
        for(int j = 0; j < 10; j++)
            arr[i][j] = i*j;

    for (int i = 0; i < 10; i++)
        for(int j = 0; j < 10; j++)
            printf("%d\n", arr[i][j]);
    free(arr);
    return 0;
}

回答3:

If extra indirection isn't a concern, you can use an array of pointers.

Edit

Here's a variation on @Platinum Azure's answer that doesn't make so many calls to malloc. Besides faster allocation, all the elements are guaranteed to be contiguous:

#define ROWS 400
#define COLS 200

int **memory = malloc(ROWS * sizeof(*memory));
int *arr = malloc(ROWS * COLS * sizeof(int));

int i;
for (i = 0; i < ROWS; ++i)
{
    memory[i] = &arr[i * COLS];
}

memory[20][10] = 3;

回答4:

In the same vein as Cogwheel's answer, here's a (somewhat dirty) trick that makes only one call to malloc():

#define ROWS 400
#define COLS 200
int** array = malloc(ROWS * sizeof(int*) + ROWS * COLS * sizeof(int));
int i;
for (i = 0; i < ROWS; ++i)
    array[i] = (int*)(array + ROWS) + (i * COLS);

This fills the first part of the buffer with pointers to each row in the immediately following, contiguous array data.

回答5:

#define ROWS 400
#define index_array_2d(a,i,j) (a)[(i)*ROWS + (j)]
...
index_array_2d( memory, 20, 10 ) = -1;
int x = index_array_2d( memory, 20, 10 );

Edit:

Arrays and pointers look very much the same, but the compiler treats them very differently. Let's see what needs to be done for an array indexing and de-referencing a pointer with offset:

1. Say we declared a static array (array on the stack is just a bit more complicated, fixed offset from a register, but essentially the same):

   static int array[10];

2. And a pointer:

   static int* pointer;

3. We then de-deference each as follows:

   x = array[i];
x = pointer[i];

The thing to note is that address of the beginning of array, as well as, address of pointer (not its contents) are fixed at link/load time. Compiler then does the following:

1. For array de-reference:
   • loads value of i,
   • adds it to the value of array, i.e. its fixed address, to form target memory address,
   • loads the value from calculated address
2. For pointer de-reference:
   • loads the value of i,
   • loads the value of pointer, i.e. the contents at its address,
   • adds two values to form the effective address
   • loads the value from calculated address.

Same happens for 2D array with additional steps of loading the second index and multiplying it by the row size (which is a constant). All this is decided at compile time, and there's no way of substituting one for the other at run-time.

Edit:

@caf here has the right solution. There's a legal way within the language to index a pointer as two-dimentional array after all.

回答6:

The compiler and runtime have no way of knowing your intended dimension capacities with only a multiplication in the malloc call.

You need to use a double pointer in order to achieve the capability of two indices. Something like this should do it:

#define ROWS 400
#define COLS 200

int **memory = malloc(ROWS * sizeof(*memory));

int i;
for (i = 0; i < ROWS; ++i)
{
    memory[i] = malloc(COLS * sizeof(*memory[i]);
}

memory[20][10] = 3;

Make sure you check all your malloc return values for NULL returns, indicating memory allocation failure.

回答7:

Working from Tim's and caf's answers, I'll leave this here for posterity:

#include <stdio.h>
#include <stdlib.h>

void Test0() {
    int                             c, i, j, n, r;
    int                             (*m)[ 3 ];

    r = 2;
    c = 3;

    m = malloc( r * c * sizeof(int) );

    for ( i = n = 0; i < r; ++i ) {
        for ( j = 0; j < c; ++j ) {
            m[ i ][ j ] = n++;
            printf( "m[ %d ][ %d ] == %d\n", i, j, m[ i ][ j ] );
        }
    }

    free( m );
}

void Test1( int r, int c ) {
    int                             i, j, n;

    int                             (*m)[ c ];

    m = malloc( r * c * sizeof(int) );

    for ( i = n = 0; i < r; ++i ) {
        for ( j = 0; j < c; ++j ) {
            m[ i ][ j ] = n++;
            printf( "m[ %d ][ %d ] == %d\n", i, j, m[ i ][ j ] );
        }
    }

    free( m );
}

void Test2( int r, int c ) {
    int                             i, j, n;

    typedef struct _M {
        int                         rows;
        int                         cols;

        int                         (*matrix)[ 0 ];
    } M;

    M *                             m;

    m = malloc( sizeof(M) + r * c * sizeof(int) );

    m->rows = r;
    m->cols = c;

    int                             (*mp)[ m->cols ] = (int (*)[ m->cols ]) &m->matrix;

    for ( i = n = 0; i < r; ++i ) {
        for ( j = 0; j < c; ++j ) {
            mp[ i ][ j ] = n++;
            printf( "m->matrix[ %d ][ %d ] == %d\n", i, j, mp[ i ][ j ] );
        }
    }

    free( m );
}

int main( int argc, const char * argv[] ) {
    int                             cols, rows;

    rows = 2;
    cols = 3;

    Test0();
    Test1( rows, cols );
    Test2( rows, cols );

    return 0;
}

回答8:

int** memory = malloc(sizeof(*memory)*400); 
for (int i=0 ; i < 400 ; i++) 
{
    memory[i] = malloc(sizeof(int)*200);
}