Address of pointer to an array
arrays, c, multidimensional-array, pointers
Solution
The addresses are the same, but the types are different.
`&s[0]` is of type `int (*)[2]`, but `s[0]` is of type `int [2]`, decaying to `int *`.
The result is that when performing arithmetic on `p`, the pattern with which it walks over the array will depend on its type. If you write `p = &s[0]` and then access `p[3][1]`, you are accessing `s[3][1]`. If on the other hand you write `int *q = s[0]`, you can only use `q` to access the first subarray of `s` e.g. `q[1]` will access `s[0][1]`.
Problem
Below is an example of two dimensional array. ``` int s[5][2] = { {0, 1}, {2, 3}, {4, 5}, {6, 7}, {8, 9} }; int (*p)[2]; ``` If I write `p = &s[0];` there is no error. But if I write `p = s[0];` there is an error, even though `&s[0]` and `s[0]` will give the same address. Please let me know why there is a differnece, even though both give the same address.