Rust type mismatch in numerical operation

• i32
• i64

fn main() {
    let x: i32 = 3;
    let y: i64 = 7;

    let z = x + 1;
    assert_eq!(z, 4);
    println!("{z}");

    let z = y + 1;
    assert_eq!(z, 8);
    println!("{z}");

    //let z = x + y;
    //println!("{z}");
}

• If we remove the i32 then this works even though the default is i32.
• That’s because Rust will infere the type of the first variable from the type of the second variable and the operation.