是否可以将元组解包为函数参数? [英] Is it possible to unpack a tuple into function arguments?

问题描述

如果我想解压一个元组并将其作为参数传递，有没有办法做到这一点:

If I want to unpack a tuple and pass it as arguments is there a way to do this:

//Does not compile
fn main() {
    let tuple = (10, Vec::new());
    foo(tuple);
}
fn foo(a: i32, b: Vec<i32>) {
    //Does stuff.
}

而不必这样做:

fn main() {
    let tuple = (10, Vec::new());
    foo(tuple.0, tuple.1);
}
fn foo(a: i32, b: Vec<i32>) {
    //Does stuff.
}

推荐答案

在夜间编译器上:

#![feature(fn_traits)]

fn main() {
    let tuple = (10, Vec::new());
    std::ops::Fn::call(&foo, tuple);
}
fn foo(a: i32, b: Vec<i32>) {
}

AFAIK 没有稳定的方法可以做到这一点.

There is AFAIK no stable way to do that.

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