B0001.md

B0001

To keep Rust rules on references (either one mutable reference or any number of immutable references) on a component, it is not possible to have two queries on the same component when one requests mutable access to it in the same system.

Erroneous code example:

use bevy::prelude::*;

#[derive(Component)]
struct Player;

#[derive(Component)]
struct Enemy;

fn move_enemies_to_player(
    mut enemies: Query<&mut Transform, With<Enemy>>,
    player: Query<&Transform, With<Player>>,
) {
    // ...
}

fn main() {
    App::new()
        .add_plugins(DefaultPlugins)
        .add_systems(Update, move_enemies_to_player)
        .run();
}

This will panic, as it's not possible to have both a mutable and an immutable query on Transform at the same time.

You have two solutions:

Solution #1: use disjoint queries using Without

As a Player entity won't be an Enemy at the same time, those two queries will actually never target the same entity. This can be encoded in the query filter with Without:

use bevy::prelude::*;

#[derive(Component)]
struct Player;

#[derive(Component)]
struct Enemy;

fn move_enemies_to_player(
    mut enemies: Query<&mut Transform, With<Enemy>>,
    player: Query<&Transform, (With<Player>, Without<Enemy>)>,
) {
    // ...
}

fn main() {
    App::new()
        .add_plugins(DefaultPlugins)
        .add_systems(Update, move_enemies_to_player)
        .run();
}

Solution #2: use a ParamSet

A ParamSet will let you have conflicting queries as a parameter, but you will still be responsible for not using them at the same time in your system.

use bevy::prelude::*;

#[derive(Component)]
struct Player;

#[derive(Component)]
struct Enemy;

fn move_enemies_to_player(
    mut transforms: ParamSet<(
        Query<&mut Transform, With<Enemy>>,
        Query<&Transform, With<Player>>,
    )>,
) {
    // ...
}

fn main() {
    App::new()
        .add_plugins(DefaultPlugins)
        .add_systems(Update, move_enemies_to_player)
        .run();
}