Why can't deduced return type be omitted here?

[jamadagni]

Int : type = {
    x : int;
    
    operator= : (out this, id: int) = { x = id; }
    
    operator+ : (this, val: int) -> _ = {
        return : Int = x + val;
    }

    value : (this) x;
}

main: () = {
    a : Int = 1;
    b := a + 2;
    std::cout << b.value() << std::endl;
}

If I omit the -> _ from the operator+, I get the error:

(7,9): error: return statement with expression must be in a function with a single anonymous return value

Why is this?

Two additional questions:

1. I hope : Int = x + val is the correct CPP2 equivalent of Int(x + val) from CPP1?
2. I hope this is the correct place to be asking questions to learn CPP2?

[jamadagni]

This seems related:

operator== : (this, that) -> bool = id_ == that.id_ ;

I am not allowed to omit -> bool = which gives me:

error: operator== must return bool

I think this should be enforced as “operator== should return a boolean” rather than “operator== should be followed by -> bool”

[MaxSagebaum]

If you just omit the -> _ from the definition it currently defaults to no return type. If you currently also remove the = then it works fine. https://cpp2.godbolt.org/z/T7McKo1cK

The error for operator== seems to be regression. Also removing the = sign gives the same error. Do you want to create a bug report for it?

I hope : Int = x + val is the correct CPP2 equivalent of Int(x + val) from CPP1?

This is ok, but Int(x+val) does also work.

[MaxSagebaum]

Missed the links:

• Shorthand syntax: https://hsutter.github.io/cppfront/cpp2/functions/#generality-note-summary-of-function-defaults
• Discussion about changes: Terser function syntax? #714 (comment)