Strict-Mode：替代argument.callee.length？

Question

Javascript：權威指南（2011）有這個例子（第186頁），它不能在嚴格模式下工作，但沒有說明如何在嚴格模式下實現它 - 我可以考慮要嘗試的東西，但我想知道最佳實踐/安全/性能 - 在嚴格模式下做這種事情的最佳方式是什麼？這裏的代碼：

// This function uses arguments.callee, so it won't work in strict mode. 
function check(args) { 
    var actual = args.length;   // The actual number of arguments 
    var expected = args.callee.length; // The expected number of arguments 
    if (actual !== expected)   // Throw an exception if they differ. 
     throw Error("Expected " + expected + "args; got " + actual); 
} 

function f(x, y, z) { 
    check(arguments); // Check that the actual # of args matches expected #. 
    return x + y + z; // Now do the rest of the function normally. 
} 

Answer 1

你可以只傳遞你檢查的函數。

function check(args, func) { 
    var actual = args.length, 
     expected = func.length; 
    if (actual !== expected) 
     throw Error("Expected " + expected + "args; got " + actual); 
} 

function f(x, y, z) { 
    check(arguments, f); 
    return x + y + z; 
} 

---

或延長Function.prototype，如果你的環境中，將允許它是......

Function.prototype.check = function (args) { 
    var actual = args.length, 
     expected = this.length; 
    if (actual !== expected) 
     throw Error("Expected " + expected + "args; got " + actual); 
} 

function f(x, y, z) { 
    f.check(arguments); 
    return x + y + z; 
} 

---

或者你可以做一個返回函數，將一個裝飾功能自動檢查...

function enforce_arg_length(_func) { 
    var expected = _func.length; 
    return function() { 
     var actual = arguments.length; 
     if (actual !== expected) 
      throw Error("Expected " + expected + "args; got " + actual); 
     return _func.apply(this, arguments); 
    }; 
} 

...並使用它像這樣...

var f = enforce_arg_length(function(x, y, z) { 
    return x + y + z; 
});