嚴格模式：替代 argument.callee.length？

Question

Javascript：權威指南 (2011) 有這個示例 (p.186)，它在嚴格模式下不起作用，但沒有說明如何在嚴格模式下實現它——我能想到要嘗試的事情，但想知道最佳實踐/security/performance——在嚴格模式下做這種事情的最好方法是什么？ 這是代碼：

// 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

您可以只傳遞您正在檢查的 function。

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;
}

---

或者擴展Function.prototype如果你在一個允許它的環境中......

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;
}

---

或者你可以制作一個裝飾器 function 返回一個 function 將自動進行檢查......

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);
    };
}

...並像這樣使用它...

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