Question

In C, why does the following result in x[1] being 2?

int a = 2, x;
...
printf("x[1] = ", &x[1])

Answer 1

It doesn’t. It results in undefined behaviour where anything can happen. You cannot access elements beyond the end of an array in a defined manner.

What’s most likely happening is that a is just “above” x on the stack, which results in x[1] having the same address as a, but it’s by no means guaranteed.

This is, of course, assuming that your printf is a typo. As it stands, it doesn’t even compile. I’m assuming it’s a typo since the question title just asks about the value of x[1] rather than the output.

To get it to work, you’d have to use something like:

printf ("x[1] = %d\n", (&x)[1]);

which also prints 2 on my system, but may do something totally different elsewhere.