Lambda演算中的JavaScript匿名函数模拟算法，结果返回`undefined`

Question

我打算使用javascript闭包来模拟lambda演算中的基本算术，以定义3 = 1 + 2像这样：

0 := λf.λx.x
1 := λf.λx.f x
2 := λf.λx.f (f x)
3 := λf.λx.f (f (f x))

它应该打印三遍hello world，现在它打印一次hello world和两次undefined。有人可以帮忙解释为什么会这样以及我的代码有什么问题吗？预先感谢。

var zero = function(f) {
  return function(x) {
    return x;
  };
};

var one = function(f) {
  return function(x) {
    return f(x);
  };
};

function add(n, m) {
  return function(f) {
    return function(x) {
      return n(f)(m(f)(x));
    };
  };
}

// test, define number two and number three in lambda calculus
var two = add(one, one);
var three = add(one, two);

// function f is to console.log
var print3times = three(value => console.log(value));
print3times("hello world")
// return:
// hello world
// undefined
// undefined

Answer 1

关键是n（f）（m（f）（x））中的m（f）（x）返回undefined，因为它是console.log。然后，它首先运行它（并打印“ hello world”），但是在其上运行n（f）时，它将打印m（f）（x）的结果：未定义。如ASDFGerte所述，如果您添加“返回值”；对它。它将正常工作。

Answer 2

您在这里。

var log = function(x) {
  console.log(x)
  return x
}

var zero = function(f) {
  return function(x) {
    return x;
  };
};

var one = function(f) {
  return function(x) {
    return f(x);
  };
};

function add(n, m) {
  return function(f) {
    return function(x) {
      return n(f)(m(f)(x));
    };
  };
}

// test, define number two and number three in lambda calculus
var two = add(one, one);
var three = add(one, two);

// function f is to console.log
var print3times = three(log);
print3times("hello world")

这是踢脚的ES6翻译：

const log = x => (console.log(x), x)

const identity = x => x

const zero = f => identity

const one = f => x => f(x)

// I took the liberty of currying add
const add = n => m => f => x => n(f)(m(f)(x))

// test, define number two and number three in lambda calculus
const two = add(one)(one)
const three = add(one)(two)

// function f is to console.log
const print3times = three(log)
print3times("hello world")