Array declaration and initialization, output array element.

Question

Enlighten fool please:

int a[3]={2,1};
printf (“%d”, a[a[1]]);

In this code is initialized an array by using a comma-separated list of values enclosed in braces. Where’s the third element?
What does it a[a[1]] mean?
Why the result of code execution is one?

Answer 1

array size can be larger than the initialized array, that's not a problem, the printf("%d", a[a[1]]);
let break it down this way
a[a[1]] - - >> will 1st check the inner part which is a[1] and this point to 1 which is also 1 with this we will now have a[1] which is also 1 again, so the output is 1

Answer 2

Just remember that you have to calculate the bracket things first so
In a[a[1]] lets write it this as b[c[1]] (remember b and c are same as 'a' i have written this just for demonstration purposes) so now in b[c[1]] you should calculate the things that are in bracket first so, c[1] = 1 (the second element in 'a') then b[c[1]] = b[1] = 1 (again the second element in 'a')

And also for the question you can always make a program and try to print it in this way you will learn more