我对clojure的循环/递归蹦床进行了广义化,以便可以与间接递归一起使用：
const trampoline = f => (...args) => {
let acc = f(...args);
while (acc && acc.type === recur) {
let [f,...args_] = acc.args;
acc = f(...args_);
}
return acc;
};
const recur = (...args) =>
({type: recur,args});
const even = n =>
n === 0
? true
: recur(odd,n - 1);
const odd = n =>
n === 0
? false
: recur(even,n - 1);
console.log(
trampoline(even) (1e5 + 1)); // false但是,我必须在通话侧明确给蹦床打电话.有没有办法像循环/递归那样再次使其隐式？
顺便说一句,这是循环/重复：
const loop = f => {
let acc = f();
while (acc && acc.type === recur)
acc = f(...acc.args);
return acc;
};
const recur = (...args) =>
({type: recur,args});
最佳答案
显然,由于您希望调用蹦床,因此不能完全跳过它.最简单的事情就是将那些经过抛光处理的调用包装在所需的API中,也许是这样的：
原文链接：https://www.f2er.com/js/531262.html// Utility code
const trampoline = f => (...args) => {
let acc = f(...args);
while (acc && acc.type === recur) {
let [f,args});
// Private implementation
const _even = n =>
n === 0
? true
: recur(_odd,n - 1);
const _odd = n =>
n === 0
? false
: recur(_even,n - 1);
// Public API
const even = trampoline(_even);
const odd = trampoline(_odd);
// Demo code
console.log(
even (1e5 + 1)); // false
console.log(
odd (1e5 + 1)); // true