Answer: 14 Difficulty: 1 Warnings: dead_code, unused_variables # Hint Hygiene in `macro_rules!` only applies to local variables. # Explanation This program prints `14` because hygiene in `macro_rules!` only applies to local variables. You can imagine hygiene as a way of assigning a color to each mention of the name of a local variable, allowing for there to be multiple distinguishable local variables in scope simultaneously with the same name. At the top of `main`, suppose we consider the name of the local variable `x` to be a purple `x`. The name of the constant `K` is just plain `K`, as constants are considered items rather than local variables (you can place items outside of a function body; you cannot place local variables outside of a function body).

let x: u8 = 1;
const K: u8 = 2;

Continuing down the body of `main`, within the declaration of the macro `m!` there are identifiers `x` and `K` being used. Since there is a local variable `x` in scope, the use of the identifier `x` within the macro body picks up the same color as the local variable `x`. There is no local variable `K` in scope so the `K` within the declaration of the macro is assigned some new color, say orange.

macro_rules! m {
    () => {
        print!("{}{}", x, K);
    };
}

Next we enter a new scope (delimited by curly braces) containing another `x` and `K`. Every new local variable always introduces a new color so let's call this `x` blue. The const again is not a local variable so no color is assigned to `K`.

{
    let x: u8 = 3;
    const K: u8 = 4;

    m!();
}

When `m!()` expands, the expanded code refers to a purple `x` and an orange `K`. The purple `x` is distinguishable from the blue `x` -- the value of the purple `x` is printed which is `1`. As for the `K`, an unhygienic (uncolored) `K` is allowed to act like any color. The second `K` is shadowing the first one. It gets picked up when looking for an orange `K` and its value is printed, which is `4`. So the output of the quiz code is `14`.