JavaScript coding questions

1. Spy decorator
   Create a decorator spy(func) that should return a wrapper that saves all calls to function in its calls property.
   Every call is saved as an array of arguments.

function work(a, b) {
  alert( a + b ); // work is an arbitrary function or method
}

work = spy(work);

work(1, 2); // 3
work(4, 5); // 9

for (let args of work.calls) {
  alert( 'call:' + args.join() ); // "call:1,2", "call:4,5"
}

Solution:

function spy(f) {
  function wrapper(...args) {
    wrapper.calls.push(args);
    return f.apply(this, arguments);
  }
  
  wrapper.calls = [];

  return wrapper; 
}

function work(a, b) {
  return a + b;
}
work = spy(work);

work(1, 2) // 3
work(2, 4) // 6

for (let args of work.calls) {
  alert( 'call:' + args.join() ); // "call:1,2", "call:2,4"
}

2. Delaying decorator
   Create a decorator delay(f, ms) that delays each call of f by ms milliseconds.

function f(x) {
  alert(x);
}

// create wrappers
let f1000 = delay(f, 1000);
let f1500 = delay(f, 1500);

f1000("test"); // shows "test" after 1000ms
f1500("test"); // shows "test" after 1500ms

Solution:

// 不用箭头函数
function delay(func, timeout) {
   function wrapper(...args) {
     let ctx= this;
     setTimeout(function() {
       func.apply(ctx, args)
     }, timeout)
   }
  return wrapper;
}

// 使用箭头函数
function delay(f, ms) {
  return function() {
    setTimeout(() => f.apply(this, arguments), ms)
  };
}

3. Debounce decorator
   The result of debounce(f, ms) decorator should be a wrapper that passes the call to f at maximum once per ms milliseconds.

let f = debounce(alert, 1000);

f(1); // runs immediately
f(2); // ignored

setTimeout( () => f(3), 100); // ignored ( only 100 ms passed )
setTimeout( () => f(4), 1100); // runs
setTimeout( () => f(5), 1500); // ignored (less than 1000 ms from the last run)

Solution:

function debounce(f, ms) {
  let isCoolingDown = false;

  return function() {
    if (isCoolingDown) {
      return 
    }
    
    f.apply(this, arguments);
    isCoolingDown = true;
    
    setTimeout(() => isCoolingDown = false, ms);
  }
}

4. Throttle decorator
   Create a “throttling” decorator throttle(f, ms) – that returns a wrapper, passing the call to f at maximum once per ms milliseconds. Those calls that fall into the “cooldown” period, are ignored.

function f(a) {
  console.log(a)
};

// f1000 passes calls to f at maximum once per 1000 ms
let f1000 = throttle(f, 1000);

f1000(1); // shows 1
f1000(2); // (throttling, 1000ms not out yet)
f1000(3); // (throttling, 1000ms not out yet)

// when 1000 ms time out...
// ...outputs 3, intermediate value 2 was ignored

Solution:

function throttle(f, ms) {
  let isThrottled = false;
  let savedArgs, savedThis;
  
  function wrapper() {
    if (isThrottled) {
      savedArgs = arguments;
      savedThis = this;
      return;
    }
    
    f.apply(this, arguments);
    isThrottled = true;
    
    setTimeout(function() {
      isThrottled = false;
      if (savedArgs) {
        wrapper.apply(savedThis, savedArgs);
        savedArgs = savedThis = null;
      }
    }, ms)
  }
  
  return wrapper;
}

5. Advanced curry implementation

function sum(a, b, c) {
  return a + b + c;
}

let curriedSum = curry(sum);

// still callable normally
alert( curriedSum(1, 2, 3) ); // 6

// get the partial with curried(1) and call it with 2 other arguments
alert( curriedSum(1)(2,3) ); // 6

// full curried form
alert( curriedSum(1)(2)(3) ); // 6

Solution:

function curry(func) {

  return function curried(...args) {
    if (args.length >= func.length) {
      return func.apply(this, args);
    } else {
      return function(...args2) {
        return curried.apply(this, args.concat(args2));
      }
    }
  };

}

6. 偏函数在登录中的应用
   user 对象被修改了。现在不是两个函数 loginOk/loginFail，现在只有一个函数 user.login(true/false)。
   以下代码中，向 askPassword 传入什么参数，使得 user.login(true) 结果是 ok，user.login(fasle) 结果是 fail？

function askPassword(ok, fail) {
  let password = prompt("Password?", '');
  if (password == "rockstar") ok();
  else fail();
}

let user = {
  name: 'John',

  login(result) {
    alert( this.name + (result ? ' logged in' : ' failed to log in') );
  }
};

askPassword(?, ?); // ? only change this line

Solution:
(1). askPassword(() => user.login(true), () => user.login(false));
(2). askPassword(user.login.bind(user, true), user.login.bind(user, false));

7. fetch 设置超时

function _fetch(fetch_promise, timeout) {
  //这是一个可以被reject的promise
  const abort_promise = new Promise((resolve, reject) => {
    setTimeout(reject, timeout, "abort promise")
  });

  //这里使用Promise.race，以最快 resolve 或 reject 的结果来传入后续绑定的回调
  const abortable_promise = Promise.race([
    fetch_promise,
    abort_promise
  ]);
  
  return abortable_promise;
}

//usage:
_fetch(fetch('//a.com/b/c'), 2000)
  .then(function(res) {
    console.log(res)
  }, function(err) {
    console.log(err);
  });