为什么此代码不起作用？

Question

i get a segmentation here. However if i declare array as int** and use malloc, it works fine.

#include <stdio.h>

void display(int **p,int numRows,int numCols) //Second Method//
{
    printf("\n");
    int i,j;
    for (i = 0; i< numRows;i++)
    {

        for (j = 0;j< numCols;j++)
        {
            printf("%d\t",p[i][j]);
        }
        printf("\n");
    }
}

int main() {

    int arr[2][2]={{1,2},{4,5}};
        display(arr,2,2);
}

PS I don't need an alternative way, just tell me why this code doesn't work.

Answer 1

arr is an array of 2 arrays of 2 int s. When it's used in an expression without being the subject of either the unary & or sizeof operators, it evaluates to a pointer to its first element.

This is a "pointer to an array of 2 int ", which is the type int (*)[2] . It is not a pointer to a pointer to int , which is the type int ** . Furthermore, these types are not compatible, and you can't pass the former to a function expecting the latter.

An int ** must be pointing at something that itself is a pointer. The pointer you are passing to display is pointing at an array, not a pointer.

Answer 2

Please don't neglect warnings

expected int **' but argument is of type 'int (*)[2]

arr is an array of 2 arrays of 2 integers

Use :

void display(int p[][2],int numRows,int numCols)

Answer 3

An int** is a pointer to a pointer, while an int [2][2] contiguously stores two arrays. That's your mistake.

With int x[2][2] , when you do x[1] , you tell the compiler to access memory by skipping one set of two elements, and the resulting expression is an array of two elements. With int** x , when you do x[1] , you tell the compiler to access memory by skipping one pointer . While you access data the same way for both, the two are simply not laid out the same way.

In memory, x[2][2] = {{1, 2}, {4, 5}} looks like this:

// 4 contiguous elements
01 00 00 00  02 00 00 00    04 00 00 00  05 00 00 00

While int** x looks like this:

// pointers to integers
aa bb cc dd  aa ab ac ad
// At [aa bb cc dd]:
    01 00 00 00  02 00 00 00
// At [aa ab ac ad]:
    04 00 00 00  05 00 00 00

Therefore, to adapt for an int** , you must create an array of int* pointers dynamically and set each entry to the lower dimension of your array.