I wrote a function

// a is 2d --- n x n matrix
void f(int** a, int n); // (1)

int main()
{
const int n=3;
int a[n][n]; // (2)

/**
Question: how do I pass the matrix a to f()?

Don't modify lines (1) and (2).

Thanks.
*/
return 0;
}

1. 野比's soln does not work, test the program below yourself:

[CODE]#include <iostream>
using namespace std;
void f(int** a, int n)
{
int i;
int j;
for(i=0; i<n; ++i)
for(j=0; j<n; ++j)
a[i][j] = i+j;
}

int main(int argc, char** argv)
{
const int n=3;
int a[n][n];
int* b = &(a[0][0]);
int** c = &b;
f(c, n);
return 0;
}[/CODE]

2. tobyliying --- I just want to understand a C++ syntax.

3. I wrote the function f(), assuming that I will pass a dynamically allocated 2d array to it. Then I changed idea --- I just want to pass a 2d array on the program stack (instead of the heap).

Your suggestion is good --- but I don't know n beforehand. Thus "pointer to array of lenght n" approach is not appropriate.

野比's way

[CODE]int* b=&a[0][0]; //a pointer to the start of the array
int** c=&b; //a pointer points to the pointer which points to the array... wow.....tongue twister... (^^!)
//now you can use the c pointer as the parameter of f() instead...[/CODE]

on my system gives access violation error. b points to the start of the 2d array, c is just the address of the b variable, so that

[CODE]c[2][2] = 1;[/CODE]
fails.

thanks 野比 and aipb2007, got your guys' idea.