I a 2D array say int a[2][3]
when we call the function say add(a);
we receive it using a pointer to an array void add(int(*p)[3])
BUT
In 1D array say int b[5]
we store address of array in a simple pointer to an integer int *p; p=b;
my question is that why don't we store b in a pointer to an array ex int(*p)[5]=b;
An array name when used as a value will decay into value equal to the pointer to its first element, with that type. This means for:
int a[2][3];
The name a will decay to &a[0], which has the type int (*)[3]. But, for:
int b[5];
The name b will decay to &b[0], which has the type int *.
However, &b is a pointer to b, which means it has the type int (*)[5]. It so happens that for an array type, its address will compare equal to the address of its first element. But, &b has a different type from &b[0].
