How to convert a two-dimensional array to a pointer of pointers

Question

Suppose you have a two-dimensional array which has been initialized like

int M[5][5] = { {1,3,4,2,5},
                {3,5,1,6,2},
                {7,5,6,2,0},
                {6,2,9,7,6},
                {1,0,8,5,2} };

This M will be passed as a parameter to a function. However the function only accepts int** as parameter, not int [][5].

My question is, how to do this conversion?

@GrijeshChauhan please read the question again. Your comment does not relate to it. — Ivaylo Strandjev
– Ivaylo Strandjev, Commented Feb 14, 2014 at 15:44
@GrijeshChauhan That won't work; p could not be passed as an int**, it would be an int***. — TypeIA
– TypeIA, Commented Feb 14, 2014 at 15:44
@mbratch: why should it accept it? a bidimensional matrix is not an array of pointers, and by no it is directly convertible to it. — Matteo Italia
– Matteo Italia, Commented Feb 14, 2014 at 15:44
possible duplicate of Cannot convert double [] [] to double ** int ** and int [][] are very different things. See my answer here. — Nicu Stiurca
– Nicu Stiurca, Commented Feb 14, 2014 at 15:45
M declared like this is an array, not a pointer. if M were allocated with mallocs, then it would work. with gcc : error: cannot convert 'int ()[5]' to 'int*' for argument '1' to 'void f(int**)' — johan d
– johan d, Commented Feb 14, 2014 at 15:46

Matteo Italia · Accepted Answer · 2014-02-14 15:56:22Z

3

An array of pointers to pointers and a bidimensional array are completely different data structures in C:

int ** normally refers to an array of pointers, each of which points to a different row (allocated elsewhere)
int [5][5] under the hood is a plain linear array where the rows are stored one after each other, and the compiler automatically performs the calculations on the index to access the right element.

So, having an int[5][5], you have to build the array of pointers by yourself, which will act as an "index" for the rows of your matrix.

int * m2[5];
for(int i=0; i<5; ++i)
    m2[i]=M[i];
f(m2);

edited Feb 14, 2014 at 15:56

answered Feb 14, 2014 at 15:46

Matteo Italia

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1 Comment

Jarod42 Over a year ago

Or even int* m2[5] = {M[0], M[1], M[2], M[3], M[4]}; f(m2);

Yakk - Adam Nevraumont · Accepted Answer · 2014-02-14 16:53:55Z

0

template<unsigned...>struct indexes {};
template<unsigned Max, unsigned...Is>struct make_indexes:make_indexes<Max-1,Max-1,Is...>{};
template<unsigned...Is>struct make_indexes<0,Is...>:indexes<Is...>{};

template< class T, std::size_t X, std::size_t Y, unsigned...Is >
std::array<T*, X> as_jagged_helper( T(&arr)[X][Y], indexes<Is...> ) {
  return {arr[Is]...};
}

template< class T, std::size_t X, std::size_t Y, unsigned...Is >
std::array<T*, X> as_jagged( T(&arr)[X][Y] ) {
  return as_jagged_helper(arr, make_indexes<X>());
}

then simply call f(&(as_jagged(M)[0]));

answered Feb 14, 2014 at 16:53

Yakk - Adam Nevraumont

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