Allocate a dynamic 2d-array by passing it to another function by reference

Question

This question builds off of two previously asked questions: C++ Passing a dynamicly allocated 2D array by reference & C - Pass by reference multidimensional array with known size

I was trying to allocate memory for a 2d-Array using the answers of these previous questions, but the memory never been allocated and I receive a BAD_ACCESS error every time a try to access the array!

This is what I have:

const int rows = 10;
const int columns = 5;

void allocate_memory(char *** maze); //prototype

int main(int argc, char ** argv) {
  char ** arr;
  allocate_memory(&arr) //pass by reference to allocate memory
  return 0;
}

void allocate_memory(char *** maze) {
  int i;
  maze =  malloc(sizeof(char *) * rows);
  for (i = 0; i < rows; ++i)
      maze[i] = malloc(sizeof(char) * columns);
}

Small point: why are you using 10 for the malloc call when you have defined rows which you use immediately after? — Weather Vane
– Weather Vane, Commented Apr 7, 2016 at 8:51
Please don't try to be a three-star programmer. Other than that technically C doesn't have "pass by reference", you can only emulate it. — Some programmer dude
– Some programmer dude, Commented Apr 7, 2016 at 8:52
As for your problem, in the function you have a pointer to the pointer to the pointer. The first indirection is because you passed a pointer to the variable, but that's only local inside your function. Have you ever wondered about the dereference operator * sometimes, and when to use it? Now might be a good idea to read up about it. — Some programmer dude
– Some programmer dude, Commented Apr 7, 2016 at 8:53
Note that maze = malloc(sizeof(char *) * 10); will only change the local copy passed to it. It won't affect the pointer you passed, which remains uninitialised. — Weather Vane
– Weather Vane, Commented Apr 7, 2016 at 8:53

haccks · Accepted Answer · 2016-04-07 09:01:35Z

3

First you should note that there is no pass by reference in C, only pass by value.

Now, you need to allocate memory for maze[0] (or *maze)

*maze =  malloc(sizeof(char *) * rows);

and then

for (i = 0; i < rows; ++i)
  (*maze)[i] = malloc(sizeof(char) * columns);

edited Apr 7, 2016 at 9:01

answered Apr 7, 2016 at 8:53

haccks

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6 Comments

Insty1956 Over a year ago

can I use *maze instead of maze[0]? is it the same?

M.M Over a year ago

@Insty1956 yes, remember to write (*maze) though when using [] after

haccks Over a year ago

@Insty1956; Yes. You can.

haccks Over a year ago

*maze[i] = malloc(sizeof(char) * columns); should be (*maze)[i] = malloc(sizeof(char) * columns);. Updated the answer.

Bob__ Over a year ago

Maybe you can also change the signature to char **allocate_memory(int rows, int columns).

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