Dynamic 2D array in typedef

Question

How do I get this to work? I am not sure how many invoices a customer will be assigned to and want to leave it dynamic. Please help.

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

typedef struct _cust{
    int id;
    int invoices[][2];
} cust;

cust account[] = {
    {1,{10,100}},
    {2,{{10,100},{20,200}}},
    {3,{{10,100},{20,200},{30,300}}}
};

int main(void) {
    printf("%d\n", account[0].invoices[0][0]);
    printf("%d\n", account[1].invoices[1][0]);
    printf("%d\n", account[2].invoices[2][0]);
    return 0;
    }

When I run this code I get following error ...

error: initialization of flexible array member in a nested context

If I give fill in a number something like this int invoices[3][2], the code runs fine.

That's not how zero-length arrays work. They can only be used with dynamic allocation (as the name "dynamic" suggests). — Kerrek SB
– Kerrek SB, Commented Apr 1, 2013 at 16:39

Johnny Mnemonic · Accepted Answer · 2013-04-01 16:59:12Z

1

You can change invoices to int** invoices and then allocate dynamically using malloc(). You can allocate a 2D array like this:

int** theArray;
theArray = (int**) malloc(arraySizeX*sizeof(int*));
for (int i = 0; i < arraySizeX; i++)
   theArray[i] = (int*) malloc(arraySizeY*sizeof(int));

edited Apr 1, 2013 at 16:59

answered Apr 1, 2013 at 16:41

Johnny Mnemonic

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5 Comments

Jens Gustedt Over a year ago

This is not an allocation of a 2D array, but of an emulation of such a thing. Also, don't cast the return of malloc, this is not C++.

Johnny Mnemonic Over a year ago

What would be the real dynamic allocation of a 2D array then?

Jens Gustedt Over a year ago

Please have a look at my answer.

Johnny Mnemonic Over a year ago

You're doing the same thing except one dimension of the array is fixed to 2.

Jens Gustedt Over a year ago

No, not at all. You are allocating an array of pointers and then a different chunk for each line. Real 2D arrays are allocated contiguously and there is no indirection. The element of a 2D array (something like array[i][j]) can be computed by arithmetic.

Jens Gustedt · Accepted Answer · 2013-04-01 17:01:38Z

1

A pointer to an array with two elements would look like this:

int (*array)[2];

but it would perhaps be easier for you to use a typedef first:

typedef int pair[2];
.
.
pair * array;

and you can allocate a large chunk to such a beast just in on go. If you have a C99 (or upwards) compatible compiler this would read like

array = malloc(sizeof(pair[n]));

answered Apr 1, 2013 at 17:01

Jens Gustedt

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