How to define some features in chain rule

[JsBlueCat]

        let mut kn;
        let mut ukn;
        let mut compute_batch_idx;
        let mut known_batch_idx;

such like

let a,b,c,d;

[daboross]

I don't understand this question. Could you describe what the problem is? Perhaps this post is incomplete?

[TomP]

I believe that this was a query about declaring multiple variables in a single let statement. To me the question implies that the author does not understand that let pattern-matches, rather than being a simple declarator as in the Basic programming language. C.f. Listing 18-3 in chapter 18 of TRPL.

[OptimisticPeach]

Yup! You can't declare multiple names in one line and not ever give them a value. However, you can do so if you give them a value eventually:

let (x, y, z);

x = 1;
y = String::from("abc");
z = SomeStruct { contents };

Playground.

However! This is just pattern matching in disguise, since let statements wouldn't work with this:

let x, y, z;

You need to make the LHS place value (I think that's the proper term?) a single value, hence we package it up in an inconspicuous tuple (...).

You can extend the declaration to include types:

let (x, y, z): (usize, String, SomeStruct);

(And then that would compile all on its own)

---

Note however, that unlike certain other languages such as C/C++/C#, you can't do something like this inside a structure declaration. IE, the following is not allowed:

struct Foo {
    (x, y, z): (usize, usize, usize),
}

Since pattern matching doesn't apply in struct declarations. It does however, in struct deconstructions:

struct Foo {
    vals: (usize, usize),
}

let Foo { vals: (x, y) } = Foo { vals: (1, 2) };

[parasyte]

Perhaps it is also worth pointing out that the mut modifier can be interspersed in the pattern as desired.

let (mut a, b, mut c) = (123, 456, "foo");

a = 42; // This is fine
b = 13; // ERROR! `b` is immutable
c = "bar"; // This is fine