Passing 2-D array as argument

Question

I am trying to pass a 2-d array to a function which accept a pointer to pointer. And I have learnt that a 2-d array is nothing a pointer to pointer(pointer to 1-D array). I

Answer 1

declaration of ‘array’ as multidimensional array must have bounds for all dimensions except the first So you have to give

array[][size] //here you must to give size for 2nd or more 

For passing the array in function , array is not a pointer to a pointer but it's pointer to an array so you write like this

fun(int (*array)[])

Here if you miss the parenthesis around (*array) then it will be an array of pointers because of precedence of operators [] has higher precedence to *

Answer 2

Why don't use std::vector instead of "raw" arrays. Advantages:
1. It can dynamically grow.
2. There is no issues about passing arguments to the function. I.e. try to call void myFuntion(int array[SIZE1][SIZE2]); with array, that has some different sizes not SIZE1 and SIZE2

Answer 3

Another templated solution would be:

template<int M, int N>
void myFunction(int array[N][M])
{
}

Answer 4

#include<iostream>
 void myFuntion(int arr[3][4]);
int main()
  {
  int array[3][4]= {{1,2,3,4},{5,6,7,8},{10,11,12,13}};
 myFuntion(array);
  return 0;
 }
   void myFuntion(int arr[3][4])
   {

   }

http://liveworkspace.org/code/0ae51e7f931c39e4f54b1ca36441de4e

Answer 5

You should at least specify the size of your second dimension.

int array[][5] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8, 9 }, { 10, 11, 12, 13 } };

There is also an error which is often repeated. To pass a 2D array as argument, you have to use the following types:

void myFuntion(int (*array)[SIZE2]);
/* or */
void myFuntion(int array[SIZE1][SIZE2]);

Answer 6

  void myFunction(int arr[][4])

you can put any number in the first [] but the compiler will ignore it. When passing a vector as parameter you must specify all dimensions but the first one.