c++ How to pass a 2D array to a function that expects a pointer

Question

When I create a 2D array, I think the variable is just a pointer to the first element of the array. So I feel I should just be able to pass it straight to a function that expects a pointer as a parameter. It seems I can do this for a one dimensional array but not for multidimensional. Can someone explain why this is?

void the_function(int* the_array){}

int main(){
   int the_array[3][3] = {{1,2,3},{1,2,3},{1,2,3}};

   the_function(the_array);
}

If I try and compile this I get the following error:

candidate function not viable: no known conversion from 'int [3][3]' to 'int *' for 1st argument
void the_function(int* the_array){}
     ^

But it works if I instead use a single dimension for the_array.

I can make it work if I put a * before the argument in my function call like this: void the_function(int* the_array){}

int main(){
   int the_array[3][3] = {{1,2,3},{1,2,3},{1,2,3}};

   the_function(*the_array);
}

But I feel like that * would be dereferencing the pointer, which I don't want to do. It should still be a pointer when passed to the function.

Any ideas?

@OlzhasZhumabek No, it will be a pointer to an array of 3 `int`, i.e. `int (*)[3]`. — Some programmer dude, Aug 18 '16 at 10:33
@JoachimPileborg, so it won't decay any longer after first one? — Incomputable, Aug 18 '16 at 10:34
Other duplicates: [Passing a pointer representing a 2D array to a function in C++](http://stackoverflow.com/q/5329107/514235) ... and ... [How to pass a 2d array through pointer in c](http://stackoverflow.com/q/14012020/514235) — iammilind, Aug 18 '16 at 10:37
There is a great answer that you should read: http://stackoverflow.com/questions/4810664/how-do-i-use-arrays-in-c — Amadeus, Aug 18 '16 at 10:37

c++ How to pass a 2D array to a function that expects a pointer

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