Error: Invalid type 'int[int]' for array subscript

Question

I want to pass a dynamically sized 2D array as a parameter to a function. I know this question has been discussed multiple times in many places but even after applying all permutations and combinations of [] and *, I haven't been able to get this to work.

#include<iostream>
#include<algorithm>
using namespace std;

int bestpath(int *A, int N, int x, int y)
{   if(x>= N || y>=N)
        return 0;
    if(x == y == N-1)
        return 0;
    int value= A[x][y]; // Error: Invalid type 'int[int]' for array subscript.
    value+= max(bestpath(A, x+1, y, N), bestpath(A, x, y+1, N));
    return value;
}

int main()
{   int T, N, i, j, k;
    cin>>T;
    for(i=0; i<T; i++)
    {   cin>>N;
        int A[N][N];
        for(j=0; j<N; j++)
        {   for(k=0; k<N; k++)
            {   cin>>A[j][k];
            }
        }

        int ans= bestpath(&A[N][N], N, 0, 0);
        cout<<ans<<endl;
    }
    return 0;
}

The error occurs in the line indicated.

If I change the function definition to-

int bestpath(int *A[], int N, int x, int y)

The error comes in the function calling: cannot convert int* to int**. I want to get my basics clear of using *A, *A[], **A, and any other way in which we pass the matrix. And is there any other, better way to do this?

A[N][N] is an int (you access the element). The address of that is a int *. — Qaz
– Qaz, Commented Dec 2, 2012 at 14:05
int A[N][N]; uses variable length arrays which are a GCC extension. — Joseph Mansfield
– Joseph Mansfield, Commented Dec 2, 2012 at 14:06
In addition to what @sftrabbit said, even if you possibly could have an array whose dimensions are specified at run-time, &A[N][N] (with A declared as int A[N][N]) is undefined behaviour. — Happy Green Kid Naps
– Happy Green Kid Naps, Commented Dec 3, 2012 at 14:38

Alok Save · Accepted Answer · 2012-12-02 14:20:53Z

1

Simply call the function as:

int ans= bestpath(A, N, 0, 0);

And change your function declaration to:

int bestpath(int A[N][N], int x, int x, int y);

You do need to move around code to get it to compile further through, here is the online sample.

answered Dec 2, 2012 at 14:20

Alok Save

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2 Comments

Chocolava Over a year ago

But I need the user to enter N.

Alok Save Over a year ago

@poonamkohli: Then use a std::vector, that is what std::vector is for. Variable length arrays are not part of the standard and your code may not be portable to some compiler which does not support them.

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