Struct with bidimensional array malloc

Question

Is it possible to malloc this struct in C?

typedef struct  {
    float a[n][M];
}myStruct;

I've tried different ways with no success.

Answer 1

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

#define N 10
#define M 15

typedef struct {
    float a[N][M];
} myStruct;

int main(void)
{
    myStruct *m;

    m = malloc(sizeof(*m));
    printf("size = %zu\n", sizeof(*m));
    free(m);

    return EXIT_SUCCESS;
}

Answer 2

Assuming n and M are compile time constants, you just want

myStruct *p = malloc (sizeof myStruct);

or

myStruct *p = malloc (sizeof *p);

If you actually meant 'how do I allocate an N x M array of a struct where n and M are not known at compile time' , the answer is:

typedef struct {
   float x;
} myStruct;

...

myStruct *p = malloc (sizeof myStruct * M * N);

then access as p[M * m + n] where 0<=m<M , 0<=n<N .

Answer 3

You need a double pointer, ie a pointer to an array of pointers, like this

typedef struct  {
    float **data;
    unsigned int rows;
    unsigned int columns;
} MyStruct;

then to malloc() it

MyStruct container;

container.rows    = SOME_INTEGER;
container.columns = SOME_OTHER_INTEGER;
container.data    = malloc(sizeof(float *) * container.rows);
if (container.data == NULL)
    stop_DoNot_ContinueFilling_the_array();
for (unsigned int i = 0 ; i < container.rows ; ++i)
    container.data[i] = malloc(sizeof(float) * container.columns);

dont forget to check container.data[i] != NULL before dereferencing and also do not forget to free() all the poitners and the pointer to array of poitners too.