pointers and arrays

Question

why does the pointer array "equivalence" not work in the following case?

void foo(int** x) {
  cout << x[0][1];
}

int main( ) {
  int a[2][2] = {{1,2},{2,3}};
  foo(a);  
}

thank you

possible duplicate of [In C, are arrays pointers or used as pointers?](http://stackoverflow.com/questions/4607128/in-c-are-arrays-pointers-or-used-as-pointers) — ildjarn, May 12 '11 at 20:48
@ildjam, no its not duplicate, there the 1D case is discussed for which the pointer array "equivalence" works — user695652, May 12 '11 at 20:51
@user695652: ...Because there's no such thing as "pointer array equivalence" and there never was. In fact, your example is a direct proof of that. — AnT stands with Russia, May 12 '11 at 20:53
@user695652 : The accepted answer there also answers your question. — ildjarn, May 12 '11 at 20:53

score 11 · Accepted Answer · answered May 12 '11 at 20:53

11

The memory model of int** and int[2][2] is different.
int a[2][2] is stored in memory as:

&a     : a[0][0]
&a + 4 : a[0][1]
&a + 8 : a[1][0]
&a + 12: a[1][1]

int** x:

&x       : addr1
&x + 4   : addr2
addr1    : x[0][0]
addr1 + 4: x[0][1]
addr2    : x[1][0]
addr2 + 4: x[1][1]

while addr1 and addr2 are just addresses in memory.
You just can't convert one to the other.

answered May 12 '11 at 20:53

Daniel

1

`&x` would be of type `int***` ... did you mean `*x` or `x[INDEX]`, where `x` is dereferenced to give the `int*` it's pointing to (which in this case is pointing to an array of type `int`)? – Jason May 12 '11 at 21:24
It's actually x instead of &x as it's the address of the 2 other addresses. – Daniel May 12 '11 at 23:03

score 4 · Answer 2 · answered May 12 '11 at 20:51

4

It doesn't work because only the first level of the multidimensional array decays to a pointer. Try this:

#include <iostream>
using std::cout;

void foo(int (*x)[2]) {
  cout << x[0][1];
}

int main( ) {
  int a[2][2] = {{1,2},{2,3}};
  foo(a);
}

answered May 12 '11 at 20:51

Robᵩ

score 0 · Answer 3 · answered May 12 '11 at 20:55

0

because the type is not int **. this right for foo function

foo(int *[2]);

type of pointer a is not int ** , exactly int* [2]..

answered May 12 '11 at 20:55

iyasar

3 Answers3