Comment by justinpombrio

Rust's traits _do_ solve the expression problem.

Each data type is a `struct`. Each operation is a trait. You `impl` each trait on each struct.

This works even if you're using a library that has declared `struct A` and `struct B` and `trait F` and `trait G`, and you want to add both a `struct C` and a `trait H`, and fill out the whole 3x3 grid without modifying the library.

The library says:

    struct A { ... }
    struct B { ... }

    trait F { ... }
    impl F for A { ... }
    impl F for B { ... }

    trait G { ... }
    impl G for A { ... }
    impl G for B { ... }

    fn some_function<T: F + G>(data: T) { ... }

Your code says:

    use library::{A, B, F, G};

    struct C { ... }
    impl F for C { ... }
    impl G for C { ... }

    trait H { ... }
    impl H for A { ... }
    impl H for B { ... }
    impl H for C { ... }

    fn another_function<T: F + G + H>(data: T);

Now `library::some_function()` can be called with an A, B, or C, even though C was defined by the user of the library. And `another_function()` can be called with A, B, or C, even though A was defined by the library.