Allocate two-dimensional array

Question

How can I allocate a bi-dimensional array using malloc? This is my current code:

typedef struct object product, *pprod;
struct object{
 int type;
 int quantity;
 pprod next;
};
pprod t[4][3];

Thanks a lot for your help.

• Daniel Fischer: What is the aim? To malloc some memory to use it as a pprod[n][3]?
• Carla Pateiro: Actually I need it to be pprod[n][m]
• Lefsler: You can create a vector (singlePinter) with the size of n*m and access with [index/rowSize][index%rowSize] so you don`t have to create and array of pointers.
• Eric Postpischil: @demonofnight: C supports variable-length arrays, since 1999, so you do not need to do manual address arithmetic.

Answer 1

To allocate memory in such a way that the layout is compatible with a normal two-dimensional array - or array of arrays - you need a pointer to an array of appropriate size,

pprod (*t)[m] = malloc(n * sizeof *t);

Thus t is a pointer to arrays of m elements of type pprod, and you can simply use it

t[i][j]

as if it were declared

pprod t[n][m];

(if malloc doesn't return NULL).

This allocation allocates a contiguous block of memory, unlike allocating a pprod ** would.

(Note: If m is not a compile-time constant, that requires that the compiler supports variable length arrays, it would not work with MSVC.)

• Carla Pateiro: That's the message i'm getting in Xcode "Initializer element is not a compile-time constant"
• Daniel Fischer: That variable length arrays are not supported? Hmm, I thought Xcode used llvm-gcc, and that should support VLAs. What's the compiler version you are using? Or are you perhaps compiling as C++? C++ doesn't have VLAs.
• Daniel Fischer: @CarlaPateiro "Initializer element is not a compile-time constant" looks like you try to initialise an array (or a struct) of static storage duration with something that is not a constant expression. Can you post the exact error message with the corresponding line of the source code?
• Carla Pateiro: The compiler version is: Apple LLVM compiler 4.2 This is what im trying to do: pprod (*t)[3] = malloc(4 * sizeof *t)
• Daniel Fischer: @CarlaPateiro At file scope? You can't call malloc at file scope, you must do that in a function.

Answer 2

For 2D array you should define a pointer like:

typedef struct obj OBJECT;
OBJECT **2Dptr = malloc (sizeof(OBJECT*)*rows)
for(i=0;i<rows;i++)
 2Dptr[i]=malloc(sizeof(OBJECT)*total_objects) //columns

There are other ways too, you can define array of pointers to your struct object.

if you want object[5][10]
you can create 5 pointers to array of 10 objects;

if you want the memory to be contiguous then you could do

*2Dptr=malloc(sizeof(OBJECT) * rows * cols); //allocate contiguosly
**access_ptr = malloc(sizeof(OBJECT*) * rows);
for(i=0;i<row;i++)
 access_ptr[i]= 2Dptr+(i*cols);

• Dineshkumar: downvote is great if downvoter lets me know what am i missing or wrong! so that everyone can learn!
• Eric Postpischil: This does not create a two-dimensional array. It creates a one-dimensional array of pointers to one-dimensional arrays. The resulting memory is generally not contiguous and cannot be passed to routines expecting two-dimensional arrays. It also causes unnecessary pointer lookups and prevents the compiler from making certain optimizations that are possible with two-dimensional arrays.
• Dineshkumar: @EricPostpischil: even if its not contiguous it can be passed to functions right?
• Eric Postpischil: It will not serve as a two-dimensional array when passed to a function expecting a two-dimensional array. It could be passed to a function expecting a pointer to an array of pointers to arrays of objects. They are not the same thing, and needless pointers-to-pointers should be avoided.
• Eric Postpischil: No, an expression of type int foo[m][n] (array of array of int) is converted to int (*)[n] (pointer to array of int). Only the first dimension is converted. int (*)[n] and int ** are not compatible.