#include <stdio.h>
int main()
{
int a [2][3][2]={{{1,2},{3,4},{5,6}},{{5,8},{9,10},{11,12}}};
printf("%d\n%d\n%d\n",a[1]-a[0],a[1][0]-a[0][0],a[1][0][0]-a[0][0][0]);
return 0;
}
The output is 3 6 4. Can anyone explain to me the reason for this? How come a[1]-a[0]=3 and a[1][0]-a[0][0]=6 and how a[] and a[][] interprets in a 3-dimensional array?
It might help if you understand how an array like yours is laid out in memory:
+------------+ Low address +---------+ Low address +------+ | a[0][0][0] | | a[0][0] | | a[0] | | a[0][0][1] | | | | | | a[0][1][0] | | a[0][1] | | | | a[0][1][1] | | | | | | a[0][2][0] | | a[0][2] | | | | a[0][2][1] | | | | | | a[1][0][0] | | a[1][0] | | a[1] | | a[1][0][1] | | | | | | a[1][1][0] | | a[1][1] | | | | a[1][1][1] | | | | | | a[1][2][0] | | a[1][2] | | | | a[1][2][1] | | | | | +------------+ High address +---------+ High address +------+
Then it helps to know that the difference you get is in multiples of the type. So for a[0] and a[1] the type is int[3][2] and there are three of those multiples between a[0] and a[1].
Same for a[0][0] and a[1][0], the type is int[2], and the difference is six int[2] units between a[0][0] and a[1][0].
To elaborate a little: Between a[0] and a[1] you have a[0][0], a[0][1] and a[0][2]. Three entries.
Between a[0][0] and a[1][0] you have a[0][0][0], a[0][0][1], a[0][1][0], a[0][1][1], a[0][2][0] anda[0][2][1]. Six entries.
a[0] decays to &a[0][0]. What the compiler is doing is subtracting the pointers for &a[0][0] from &a[1][0], and dividing that difference with the size of the type (which is int[3][2]).At the point of address, a[1] and a[1][0] are the same value. And a[0] and a[0][0] are same value.
But the types are different.
a[1][0] and a[0][0] are int *, from a[0][0] to a[1][0], there are 6 int.
And from a[1] to a[0], there are 3 {x, y}.
a[1][0][0] and a[0][0][0] are int, a[1][0][0]-a[0][0][0] = 5 - 1 = 4.
In C, a multi-dimensional array is conceptually an array whose elements are also arrays. So if you do:
int array[2][3]; Conceptually you end up with:
array[0] => [0, 1, 2]
array[1] => [0, 1, 2]
int array[2][3][2]; ...will give you a structure like:
array[0] => [0] => [1, 2]
[1] => [3, 4]
[2] => [5, 6]
array[1] => [0] => [5, 8]
[1] => [9, 10]
[2] => [11, 12]
a[1]-a[0] => will give difference you get is type of unit. a[0] and a[1] is int and there are three unit between them.similarly for the second part
a[1][0]-a[0][0]=6
number of combination for between a[0][0] and a[1][0] is 6.
