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Pointer of a 2D array in C++?

开发者 https://www.devze.com 2023-01-11 16:06 出处:网络
I am writing a function to rotate a NxN matrix by 90 degree. Here is my function // \"matrix\" is an n by n matrix

I am writing a function to rotate a NxN matrix by 90 degree.

Here is my function

// "matrix" is an n by n matrix  
void rotateMatrix(int ** matrix, int n)
{
    for(int layer=0; layer < n; layer++)
    {
        int first = layer, last = n - layer -1;
        for(int i=0; i<n; i++)
      开发者_如何学JAVA  {
            int temp = matrix[i][last];
            matrix[i][last] = matrix[first][i];
            matrix[first][i] = matrix[i][first];
            matrix[i][first] = matrix[last][i];
            matrix[last][i]=temp;
        }
    }
}

Here is how I initialize and call the function in main function:

int m[5][5] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25};
rotateMatrix(m,5);

What I got from my IDE is:

> error: cannot convert ‘int (*)[5]’ to
> ‘int**’ for argument ‘1’ to ‘void
> rotateMatrix(int**, int)’

I kinda know why it's wrong since "m" is a int* . However, I am not sure I can I solve this problem?


Are the dimensions of the matrix known at compile time? If so:

template <size_t n>
void rotateMatrix(int (&matrix)[n][n])
{
    ...
}


To actually solve your problem, assuming you're sticking with plain ol' arrays, and that the size needs to be arbitrary, you're going to have to make a couple of changes.

First, you need to pass in just a single pointer int*. A 2D array is not an array of pointers (or a pointer to a pointer), it's just a 1D array that the compiler knows to treat as a 2D array (i.e. the [3][4] element in a 5x5 array is at position 3*5+4). This does mean that you'll have to typecast the 2D array before passing it in, since the compiler doesn't want to discard that extra information.

Second, you'll have to access it as a 1D array. Like the example above, m[i][j] can be found at m[n*i+j].

Edit, to address the downvotes: this answer is not intended as a full explanation of array and pointer typing. It is simply trying to tell the OP a way to pass an arbitrary-sized 2D array into a function and perform some operation on it. I don't believe I said anything incorrect, just incomplete.


An array is not exactly the same thing as a pointer. In particular, a pointer to an array is not a pointer to a pointer. It's just a regular pointer to the first object of the array. An array of arrays likewise doesn't contain any pointers.

Much array confusion can be alleviated in C++ by using array references:

void rotateMatrix(int (&matrix)[5][5], int n) // matrix is ref to 5x5 array

Now there is no pointer vs array confusion whatsoever.


You can keep your function the same, but you cannot pass static arrays in. You'll need to make dynamic arrays.

int size = 5;
int** m; 
m = new int**[size];
for (int i =0; i < size; ++i){
   m[i] = new int*[size];
}

/* Assign values to every element of m, using a for loop */
rotateMatrix(m, size);
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