Can you explain this in more detail? I'm not really familiar with Mojo but I can't think of obvious ways in which Rust's semantics (as opposed to syntax) are more functional than C++'s.
Rust's type system is very Haskell/ML inspired, specifically the first class support for algebraic types and pattern matching.
Modern C++ has come a long way with typed optionals and variants, but it's very tacked-on and lacks the safety guarantees and ergonomics of a language designed with this paradigm at a fundamental level.
So what does this actually translate to in terms of Mojo? Rust has algebraic types and pattern matching, C++ has some specific use cases that are "tacked-on", to use your phrasing, so when the parent comment says that Mojo is what a good C++ would look like, they're saying...it doesn't have algebraic types, or that they're tacked on? It's not clear to me why that would be a selling point, so giving the parent commenter the benefit of the doubt, I don't think this sounds like what they were trying to say.
I don't know about Mojo. I was just answering "...I can't think of obvious ways in which Rust's semantics (as opposed to syntax) are more functional than C++'s"
I consider it basically an accident of history that the first non-research languages to support ADTs were functional. (The first language to have them was actually CLU, which was imperative, but it was a research language.) There isn't really anything about ADTs that makes them incompatible with imperative languages.
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Can you explain this in more detail? I'm not really familiar with Mojo but I can't think of obvious ways in which Rust's semantics (as opposed to syntax) are more functional than C++'s.
Rust's type system is very Haskell/ML inspired, specifically the first class support for algebraic types and pattern matching.
Modern C++ has come a long way with typed optionals and variants, but it's very tacked-on and lacks the safety guarantees and ergonomics of a language designed with this paradigm at a fundamental level.
So what does this actually translate to in terms of Mojo? Rust has algebraic types and pattern matching, C++ has some specific use cases that are "tacked-on", to use your phrasing, so when the parent comment says that Mojo is what a good C++ would look like, they're saying...it doesn't have algebraic types, or that they're tacked on? It's not clear to me why that would be a selling point, so giving the parent commenter the benefit of the doubt, I don't think this sounds like what they were trying to say.
I don't know about Mojo. I was just answering "...I can't think of obvious ways in which Rust's semantics (as opposed to syntax) are more functional than C++'s"
I consider it basically an accident of history that the first non-research languages to support ADTs were functional. (The first language to have them was actually CLU, which was imperative, but it was a research language.) There isn't really anything about ADTs that makes them incompatible with imperative languages.