Lib.rs

mex a parser/evaluator for simple mathematical expressions, with no dependencies at all. Currently, it supports only a tiny amount of syntax. Here are some valid expressions:

>> 10
10

>> 10 - 3 * 10
-20

>> π = 3.1415926536
3.1415926536

>> 5 * π
15.707963268

And that's it, for now.

It supports the following operators: +, -, *, /, =.

But it's easy to add more (by editing the source).

The crate can be used as a binary or a library. The binary is a REPL, where you enter single expressions line by line and get the result given back to you. An example of the library is below:

use mex::parser;
use mex::evaluator;
use mex::evaluator::context;
use mex::evaluator::object;

fn main() {
    let input = "2 * x";
    let mut parse = parser::Parser::new(input);

    let mut context = context::Context::new();
    context.set(&String::from("x"), object::Object::Number(13.5));

    match parse.parse() {
        Ok(node) => {
            match evaluator::eval_node_in(node, &mut context) {
                Ok(result) => println!("{}", result),
                Err(err) => println!("eval error: {:?}", err),
            }
        }

        Err(e) => println!("parse error: {:?}", e),
    };
}

In the future, I'd like to add these things:

• Simple API for simple uses

  • e.g. mex::evaluate("1 + 2") → 3

• Functions

  • e.g. f(x, y) = x + y
  • e.g. f(2, 3) → 5

• Customisation / options

  • Probably via bitflags
  • Things like which operators are allowed, if you're allowed to define variables, etc...

• Builtins

  • e.g. π, e, sin(θ)
  • Going back to the last point: builtins should be optional

• Multiple expressions in one

  • e.g. 5 + {1, 2, 3} → 6, 7, and 8
  • They would be returned in a hash set, probably

• Symbolic Expressions

  • e.g. 'x = a + b' (maybe with different delimiters)
  • Can be either equations (i.e. with equals), or just simple expressions
  • where clause
    • e.g. 'a + b' where a = 2, b = 10 → 12
  • solve ... for
    • e.g. solve 'a = x + b' for x → 'x = a - b'
    • e.g. (solve 'a = x + b' for x) where a = 5, b = 2 → 3
    • e.g. solve 'x^2 + 3x + 2' for x → (x + 1)(x + 2)
  • Maybe even integration/differentiation