is a==*a?? A query regarding pointers
arrays, c, pointers
Solution
Arrays are converted to pointer when used in an expression except when they are an operand of `sizeof` and unary `&` operator. `a` and `*a` are of different types (after decay) but have the same address value.
- `a` decays to pointer to first element (first row) of array and is of type `int (*)[3]`.
- `*a` dereference the row pointed by `a` and further decayed to pointer to first element of first row. It is of type `int *`.
- `a[0]` is representing the first row which is of type `int [3]`. In expression it decays to pointer to first element of first row and is of type `int *` after decay.
As the address of an array the address of first byte, therefore address of an array, address of first row and address of first element all have the same value. So, after decay, all of `a`, `*a` and `a[0]` points to same location.
Here is a graphical view of the above explanation:
What exactly is the array name in c?
Problem
``` int main() { int a[4][3] = {10,20,30,40,50,60,70,80,90,100,110,120}; printf("%d",((a==*a) && (*a==a[0]))); return 0; } ``` Prints 1 on the console. Anyone has logical explanation??