可选回调的 JavaScript 样式

I personally prefer

typeof callback === 'function' && callback();

The typeof command is dodgy however and should only be used for "undefined" and "function"

The problems with the typeof !== undefined is that the user might pass in a value that is defined and not a function

You can also do:

var noop = function(){}; // do nothing.


function save (callback){
callback = callback || noop;
.....do stuff......
};

It's specially useful if you happen to use the callback in a few places.

Additionally if you are using jQuery, you already have a function like that, it's called $.noop

If the criteria for running the callback is that whether its defined or not, then you're fine. Also, I suggest to check if its really a function in addition.

I got so tired of seeing that same snippet over and over I wrote this:

  var cb = function(g) {
if (g) {
var args = Array.prototype.slice.call(arguments);
args.shift();
g.apply(null, args);
}
};

I've got hundred of functions doing things like

  cb(callback, { error : null }, [0, 3, 5], true);

or whatever...

I'm skeptical of the whole "make sure it's function" strategy. The only legitimate values are a function or falsy. If someone passes in a non-zero number or a non-empty string, what are you going to do? How does ignoring the problem solve it?

I have sinced moved to coffee-script and found default arguments is a nice way to solve this problem

doSomething = (arg1, arg2, callback = ()->)->
callback()

It can easilly be done with ArgueJS:

function save (){
arguments = __({callback: [Function]})
.....do stuff......
if(arguments.callback){
callback();
};
};

Simply do

if (callback) callback();

I prefer to call the callback if supplied, no matter what type it is. Don't let it fail silently, so the implementor knows he passed in an incorrect argument and can fix it.

A valid function is based on the Function prototype, use:

if (callback instanceof Function)

to be sure the callback is a function

ECMAScript 6

// @param callback Default value is a noop fn.
const save = (callback = () => {}) => {
callback(); // Executes callback when specified otherwise does nothing
};

Rather than make the callback optional, just assign a default and call it no matter what

const identity = x =>
x


const save (..., callback = identity) {
// ...
return callback (...)
}

When used

save (...)              // callback has no effect
save (..., console.log) // console.log is used as callback

Such a style is called continuation-passing style. Here's a real example, combinations, that generates all possible combinations of an Array input

const identity = x =>
x


const None =
Symbol ()


const combinations = ([ x = None, ...rest ], callback = identity) =>
x === None
? callback ([[]])
: combinations
( rest
, combs =>
callback (combs .concat (combs .map (c => [ x, ...c ])))
)


console.log (combinations (['A', 'B', 'C']))
// [ []
// , [ 'C' ]
// , [ 'B' ]
// , [ 'B', 'C' ]
// , [ 'A' ]
// , [ 'A', 'C' ]
// , [ 'A', 'B' ]
// , [ 'A', 'B', 'C' ]
// ]

Because combinations is defined in continuation-passing style, the above call is effectively the same

combinations (['A', 'B', 'C'], console.log)
// [ []
// , [ 'C' ]
// , [ 'B' ]
// , [ 'B', 'C' ]
// , [ 'A' ]
// , [ 'A', 'C' ]
// , [ 'A', 'B' ]
// , [ 'A', 'B', 'C' ]
// ]

We can also pass a custom continuation that does something else with the result

console.log (combinations (['A', 'B', 'C'], combs => combs.length))
// 8
// (8 total combinations)

Continuation-passing style can be used with surprisingly elegant results

const first = (x, y) =>
x


const fibonacci = (n, callback = first) =>
n === 0
? callback (0, 1)
: fibonacci
( n - 1
, (a, b) => callback (b, a + b)
)
        

console.log (fibonacci (10)) // 55
// 55 is the 10th fibonacci number
// (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...)