5 ms·

I have the same problem with almost all new languages including Rust: everything should be an expression and not a statement. Even conditional logic constructs

by agentgt 10y ago

I have the same problem with almost all new languages including Rust: everything should be an expression and not a statement. Even conditional logic constructs (if/else) should be an expression (i.e. return a value). The language doesn't even have to be pure or hard core FP to follow this philosophy.

nathancahill 10y ago
I highly agree, especially about the if/else. The more I code, the more I try to elimante branching. If statements as expressions would be ideal.
- qznc 10y ago
  Most mainstream languages (C, Java, etc) have this. The ternary operator ?:
  - 35bge57dtjku 10y ago
    That's like farting in the bathtub and calling it a hot tub.
    - tigershark 10y ago
      Why? It is an expression, not a statement regardless of your farting nonsense.
      - 35bge57dtjku 10y ago
        Because they aren't the same thing.
hermanradtke 10y ago
Based on my understanding, Rust is an expressive language. What have you seen that is not expressive?
captainmuon 10y ago
Actually, in Nim almost everything is an expression, including if blocks, case, and even except blocks. Together with immutable "variables" (let) this encourages a pretty functional style. I'd just wish their support for lambdas and list comprehensions would be a bit better.
jeremyjh 10y ago
You are mistaken about rust. I can't think of a construct that is not an expression.
- steveklabnik 10y ago
  let x = 5; (That said, very nearly everything is an expression, yes)
  - wisam 10y ago
    I don't know about Rust; but in ML family, everything in top level is a binding. Whether it's a variable, function, type or structure. SML is explicit about it and its syntax makes it clear what type of binding is going on. val x = 5 fun f = fn x => x * x datatype weekend = Sat | Sun So while on top level not everything is an expression, everything on the RHS is an expression.
    - qwertyuiop924 10y ago
      In rust, assignment isn't an expression, but everything else is.

by agentgt 10y ago

nathancahill 10y ago

I highly agree, especially about the if/else. The more I code, the more I try to elimante branching. If statements as expressions would be ideal.

qznc 10y ago
Most mainstream languages (C, Java, etc) have this. The ternary operator ?:
- 35bge57dtjku 10y ago
  That's like farting in the bathtub and calling it a hot tub.
  - tigershark 10y ago
    Why? It is an expression, not a statement regardless of your farting nonsense.
    - 35bge57dtjku 10y ago
      Because they aren't the same thing.

hermanradtke 10y ago

Based on my understanding, Rust is an expressive language. What have you seen that is not expressive?

captainmuon 10y ago

Actually, in Nim almost everything is an expression, including if blocks, case, and even except blocks. Together with immutable "variables" (let) this encourages a pretty functional style. I'd just wish their support for lambdas and list comprehensions would be a bit better.

jeremyjh 10y ago

You are mistaken about rust. I can't think of a construct that is not an expression.

steveklabnik 10y ago
let x = 5; (That said, very nearly everything is an expression, yes)
- wisam 10y ago
  I don't know about Rust; but in ML family, everything in top level is a binding. Whether it's a variable, function, type or structure. SML is explicit about it and its syntax makes it clear what type of binding is going on. val x = 5 fun f = fn x => x * x datatype weekend = Sat | Sun So while on top level not everything is an expression, everything on the RHS is an expression.
  - qwertyuiop924 10y ago
    In rust, assignment isn't an expression, but everything else is.