跳过JavaScript函数中的参数

问题描述

我有这样的功能:

function foo(a, b, c, d, e, f) {
}

为了仅使用 f 参数调用此函数，我知道我应该这样做:

In order to call this function only with an f argument, I know I should do:

foo(undefined, undefined, undefined, undefined, undefined, theFValue);

有没有那么冗长的方法呢?

Is there a less verbose way to do this?

解决方案:
我选择了一些建议的解决方案(不使用辅助函数)

Solutions:
I selected some proposed solutions (without using helper third functions)

// zero - ideal one, actually not possible(?!)
foo(f: fValue);

// one - asks a "strange" declaration
var _ = undefined;
foo(_, _, _, _, _, fValue);

// two - asks the {} to be used instead of a 'natural' list of args
//     - users should be aware about the internal structure of args obj
//       so this option is not 'intellisense friendly'
function foo(args){
    // do stuff with `args.a`, `args.b`, etc.
}
foo({f: fValue});

推荐答案

如此:

foo(undefined, undefined, undefined, undefined, undefined, arg1, arg2);

.等于:

foo(...Array(5), arg1, arg2);

.or:

foo(...[,,,,,], arg1, arg2);

---

这样:

foo(undefined, arg1, arg2);

.等于:

foo(...Array(1), arg1, arg2);

.or:

foo(...[,], arg1, arg2);

---

这样:

foo(arg1, arg2);

.等于:

foo(...Array(0), arg1, arg2);

.or:

foo(...[], arg1, arg2);

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