Passing 2D array in C

Question

I want to display the elements of a 2D array by passing a pointer to a function. I successfully did it for 1D array.

#include<stdio.h>

void displaymat(int *a);

int main()
{
  int a[3]={0,1,2};
  int t[3][3]={1,2,3,4,5,6,7,8,9};

  displaymat(a);

  return 0;
}

void displaymat(int *a)
{
  int i;

  for(i=0;i<3;i++)
    printf("%d\n",a[i]);/*works for single dimensional array*/
}

But when I use displaymat(t), it gives me an error saying incompatible pointer type. However displaymat(&t[0][0]) seems to work. Why this apparent difference in passing pointer between 1D and 2D arrays?

int t[3][3]={1,2,3,4,5,6,7,8,9}; is a 1-D array. The value of t[1][2] is dependent on whether data is stored in row-major or column-major order. Are you sure this is what you intended to do? — IceArdor
– IceArdor, Commented Dec 4, 2013 at 18:14
@Iceardor-My bad, with proper initialization of the 2D array, displaymat(t) seems to work. What exactly is int t[3][3]={1,2,3,4,5,6,7,8,9}; then and why displaymat(t) gives me an error? — Vaibhav Sundriyal
– Vaibhav Sundriyal, Commented Dec 4, 2013 at 18:25

Dr.Tower · Accepted Answer · 2013-12-04 18:24:31Z

2

t is of type int**, which is a pointer to an array of int pointers.

a is of type int*

&t[0][0] is of type int*.

You should be able to call displaymat(t[0]), displaymat(t[1]), and displaymat(t[2])

answered Dec 4, 2013 at 18:24

Dr.Tower

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