如何使闭包键入“ extern“ C” fn”

Question

This is short example about this question.

#[fixed_stack_segment]
fn test(func: extern "C" fn() -> ~str) -> ~str {
    func()
}
extern "C" fn func1() -> ~str {
    ~"hello"
}

fn main() {
    let func2 = || -> ~str { ~"world" };
    println(test(func1));
    println(test(func2));
}

Then, rustc stops with error.

st.rs:13:17: 13:22 error: mismatched types: expected `extern "C" fn() -> ~str` but found `&fn<no-bounds>() -> ~str` (expected extern fn but found fn)
st.rs:13     println(test(func2));

I cannot find a way to make the lambda be an extern fn.

What should I do?

Answer 1

The closure syntax is always either &fn or ~fn , and to make an extern "ABI" fn (for any value of ABI , including Rust ), one needs to use a full function declaration.

#[fixed_stack_segment]
fn test(func: extern "C" fn() -> ~str) -> ~str {
    func()
}
extern "C" fn func1() -> ~str {
    ~"hello"
}

fn main() {
    extern "C" fn func2() -> ~str { ~"world" } 

    println(test(func1));
    println(test(func2));
}

There is some talk of allowing lambdas to create non-closures too, with the ABI inferred like anything else in a type signature, but this isn't implemented yet.

---

However, as Vladimir Matveev says, there is a fundamental difference between closures and normal functions, which means that one will never be able to use all the features of closures when passed as an extern fn . The difference is closures can capture (references to) variables, ie

let n = 1;
let f = || { n + 1 }; 

This means a closure is effectively represented by

struct AndFn { // &fn
    env: &Environment,
    func: extern "Rust" fn()
}

struct TwiddleFn { // ~fn
    env: ~Environment,
    func: extern "Rust" fn()
}

where Environment is a struct that contains all the capture variables for that closure (it differs for each &fn , since each one captures different things); the func is a function pointer to the code of the closure, which is what gets executed when the closure is called; if a closure captures any variables, the func will require the env to exist. Thus, the lambda syntax will only be able to create plain extern fn s when it captures no variables.