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;
}

你到底是要干什么,
请说具体点
做这个矩阵 是干什么用。
具体有什么作用呀

I suppose that you wanna pass and use the matrix inside the f() function...
I made it like this:
--------------
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...

here's my test result..

TO HJin：
你想用指想指针的指针指向一个2d数组似乎不能实现。因为使用数组名，将指向数组的第一个元素，这里2维数组即是指向一个数组，然后你想进一步去实现这个子数组的名字转换，似乎是不允许的，至少我没看到过。

如果你想用指针去指向一个2维数组，可以这样：
const int n=3;
int a[n][n]; // (2)
int (*p)[n] = a;

这样一来，你的函数参数就要边一下了，
void f(int (*p)[3], int n); // (1)

正如你所见，数组第2维大小必须固定。


我只说了我知道了，也许有其他更好的方法，期待！

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.

what about this??
make a pointer array(1-d)..

int* b[n];

then.. store each address of the 2-d array rows to b[n]..

for(int i=0;i<n;++i){
b[i]=&a[i][0]; //suppose the 2-d array is a[n][n]
}

now you can use b as parameter of f() directly...
(b is a pointer to a pointer array, so it is in the form int** , just like argv in "char** argv")

野比's way is OK!

看到那个char** argv我突然明白HJin想做什么了...
Am I right, HJin?

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.