Addresses of 1d arrays are actually taken as
a[i]=*(a+i);
Are the addresses of 2d arrays calculated as
a[i][j]=**(a+i+j);
A:
No, because then a[1][2] and a[2][1] would be at the same place. Something like *(a+i*n+j) for an n-by-m array is closer to the mark (though beware I type the exact expression into a markdown editor, not a unit test).
Graham Lee
2010-09-27 11:02:11
A:
// CTest.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
void print (int i, int j )
{
int a[3][3] = {
{ 1,2,3 },
{ 4,5,6 },
{ 7,8,9 }
};
printf ("%d\n", *(*(a+i)+j) );
printf ("%d\n", a[i][j] );
}
int _tmain(int argc, _TCHAR* argv[])
{
print (0,0);
print (1,1);
print (2,2);
return 0;
}
Returns:
1 1 5 5 9 9
*This has been run through a compiler....
Tony Lambert
2010-09-27 11:03:04
I am sorry but its not working
Shweta
2010-09-27 11:08:19
This is supposed to work. What exactly doesn't work?
codymanix
2010-09-27 11:14:45
@codymanix: There is no double dereferencing, so the type of all that is pointer-to-int.
Oli Charlesworth
2010-09-27 11:17:24
This definitely works now I tried it...
Tony Lambert
2010-09-27 15:34:31
+2
A:
The other answers are not quite correct. It's more like:
*(*(a+i)+j)
Oli Charlesworth
2010-09-27 11:09:23
A:
No. a and a[i] are of different types (respectively int** and int*).
Supposing that in your example a was defined as array of array of int (eg a[10][20]), when you pass it to a function (thereby converting it to pointer to the first element of the array) you have (with further "simplifications" for the 2nd array rank)
a is of type `int**`
a[i] is of type `int*`
a[i][j] is of type `int`
*(a+i) is of type `int*`
a+i+j is of type `int**`
*(a+i+j) is of type `int*`
*(*(a+i)+j) is of type `int`
pmg
2010-09-27 11:15:10
If `a` is indeed a "2d array" (by which we normally mean an array of arrays) then it decays to type `int (*)[n]`.
caf
2010-09-27 11:25:49
Yes, that's why I wrote "further simplification for the 2nd rank". Thank you for making it clearer.
pmg
2010-09-27 11:28:17
It's not right though - if `a` is defined as `a[10][20]` then `a` is **not** of type `int**`, it's of type `int(*)[20]`.
caf
2010-09-28 11:29:33
A:
Apply the rule recursively:
a[i][j] == *(a[i] + j) == *(*(a + i) + j)
John Bode
2010-09-27 12:15:27