Answer: 54
Difficulty: 1

# Hint

`..` means one thing in an expression and something else in a pattern.

# Explanation

This question demonstrates two different meanings of `..`.

In expression position, `..` is the syntax for constructing various types of
ranges. Here the expression `(0, 1, ..)` is a tuple with three elements, the
third one having type [`RangeFull`].

[`RangeFull`]: https://doc.rust-lang.org/std/ops/struct.RangeFull.html

On the other hand in a pattern, `..` is used to mean "any number of elements".
So the pattern `(.., x, y)` matches a tuple with 2 or more elements, binding the
second-last one to `x` and the last one to `y`.

Coming out of the first line of `main`, we have `x = 1` and `y = (..)`. Thus the
value printed is going to be `b"066"[..][1]`.

The expression `b"066"` is a byte-string literal of type `&'static [u8; 3]`
containing the three ASCII bytes `b'0'`, `b'6'`, `b'6'`.

When we slice the byte-string with `RangeFull` we get a dynamically sized slice
`[u8]` of length 3. Next we access element `1` of the slice, which is the byte
`b'6'` of type `u8`. When printed, we see the decimal representation of the byte
value of the ASCII digit 6, which is the number 54.