Can’t understand this way to calculate the square of a number

Obviously a hack… but a way of squaring a number without using the * operator (this was a coding contest requirement).

(&a)[n] 

is equivalent to a pointer to int at location

(a + sizeof(a[n])*n)

and thus the entire expression is

  (&a)[n] -a 

= (a + sizeof(a[n])*n -a) /sizeof(int)

= sizeof(a[n])*n / sizeof(int)
= sizeof(int) * n * n / sizeof(int)
= n * n