If I have a function
foo()
that I call with no arguments most of the time, but one argument in special cases, is
var arg1 = arguments[0];
if (arg1) {
<special case code>
}
inside the function a completely safe thing to do?
Yes it is safe. Unless you pass in false, "", 0, null or undefined as an argument. It's better to check againts the value of undefined. (If you pass in undefined then tough! that's not a valid argument).
There are 3 popular checks
foo === undefined : Standard check but someone (evil) might do window.undefined = true
typeof foo !== "undefined" : Checks for type and is safe.
foo === void 0 : void 0 returns the real undefined
But this is prefered
function myFunction(foo) {
if (foo !== undefined) {
...
} else {
...
}
}
Yes, that's fine. A reasonable alternative is to name the argument, and not use the arguments object:
function foo(specialArg)
{
if (specialArg)
{
// special case code
}
}
Note that if(bar) tests the truthiness of bar. If you call foo with any falsy value, such asfoo(0), foo(false), foo(null), etc., the special case code will not execute in the above function (or your original function, for that matter). You can change the test to
if (typeof specialArg !=== 'undefined')
{
// ...
}
to make sure that the special case code is executed when the argument is supplied but falsy.
You can do this:
function foo(arg1){
if (arg1){
// Special case
}
else{
// No argument
}
// Rest of function
}
As long as you document the behaviour sufficiently I don't see anything wrong with it.
However you'd be better off checking the argument length, as opposed to how you're doing it now. Say for example you called:
myFunction(0);
It will never process the argument.
If it's a single optional argument you may be better off having it as a named argument in the function and checking if a defined value was passed in, depends on your use case.
The basic fact you are interested in is "was foo called with 0 or 1 argument(s)?". So I would test arguments.length to avoid future problems with a special argument that evaluates to false.
Related
From this question, given that I don't want to specify any context, hence, passing null to thisArg in call().
What would be the difference between line 2 and line 3 in the following code? Is there any benefit from doing one over the other?
function sum(a,b) { return a + b; }
var result1 = sum.call(null,3,4); // 7
var result2 = sum(3,4); // 7
Similarly for apply():
var arr = [1,2,4];
var result3 = Math.max.apply(null, arr); // 4
var result4 = Math.max(...arr); // 4
It depends on whether the function you're calling was defined in loose mode or strict mode.
When calling a loose-mode function, the following three things all call the function such that this within the call is the global object for the environment (globalThis, aka window on browsers):
Calling the function without setting this: fn()
Calling the function providing a this value of undefined: fn.call(undefined) and similar
Calling the function providing a this value of null: fn.call(null) and similar
With a strict mode function, the first two both cause this during the call to be undefined, and the third (explicitly setting it to null) sets it to (you guessed it) null.
Examples:
function loose() {
console.log(`loose: ${this === null ? "null" : typeof this}`);
}
loose();
loose.call(undefined);
loose.call(null);
function strict() {
"use strict";
console.log(`strict: ${this === null ? "null" : typeof this}`);
}
strict();
strict.call(undefined);
strict.call(null);
In the normal case, if you don't need to set this to anything in particular, just call the function so the default behavior takes place.
One wrinkle: If you have an array of arguments you need to spread out as discrete arguments to the function, in any even vaguely up-to-date environment, you can use spread notation to do that: fn(...theArray). If you're stuck in an obsolete environment, the rough equivalent is fn.apply(undefined, theArray).
If you have a specific need to set a specific this value, you can do that via call or apply (as you've found), including (for strict mode functions) undefined or null.
TJ Crowder has the detailed response here, but as a general rule:
fn.call(null): In almost all cases, just call the function.
fn.apply(null, args): This is useful in some cases where you have an array of arguments and your environment doesn't support ...args, but otherwise spreading the arguments is probably more conventional: fn(...args)
I realize that this question is extremely similar to many others, but I must be missing some nuance with checking whether a variable is set or not. I have seen this in other developers code:
if(foo){
doSomething(foo);
}
else{
alert('error of whatever');
}
With the intent that the doSomething() will only execute if foo is set or not undefined. However, when I google this, it seems everyone says "typeof" should be used instead of the above method. I specifically see this in use with angular, like this:
if($scope.property){
dothis();
}
Am I missing something? When I see the above code, it seems to work, but all the answers I see never say this is the correct way to check if something is set or exists.
For if() checks, in MOST scenarios where you are checking for the existence of a property on an object (your situation), what you have described is perfectly valid, easy and convenient. It checks for the existence of the property and returns true if it exists.
However, there are plenty of nuanced areas where a typeof check is "more" correct, particularly if your type is being coerced in any way via == or if you want to differentiate between null and undefined.
For instance, if null is a valid value for your property to have but undefined is not, in your example dothis() would still be called. You would prevent this with a typeof check.
if (typeof $scope.property === 'undefined') {
dothis();
}
Finally, if you are checking for the existence of a variable instead of the existence of a property, an exception will be thrown if the variable you are checking is not defined, forcing you to use a typeof check.
In those scenarios, verbosity is your friend.
This has to do with the concept of "truthiness". Any value besides false, 0, "", null, undefined, and NaN is "truthy" which means the first block of the if-statement will run. For instance:
if ("") {
alert("falsie"); // won't run because the empty string ("") is falsie
} else {
alert("truthie"); // will run
}
whereas
if ("something") {
alert("truthy"); // will run because "something" is truthy
} else {
alert("falsie"); // won't run
}
Going back to your example, if foo is truthy (meaning that it has ANY value other than false, 0, "", null, undefined, and NaN) then it will run the first block of the if-statement (which has the doSomething() function in it).
You could also use short circuit evaluation and do this in one line.
($scope.property && doThis())
It is better to use typeof if you need to explicitly check whether the value is undefined or not. If you just want to check if the value is truthy, then you don't need typeof.
The reason the typeof operator is preferred is because it doesn't throw a ReferenceError exception when used with an undeclared variable.
However, it is important to note that variables initialized as null will return "object", so to avoid this issue, the following code is recommended:
if (typeof variable === 'undefined' || variable === null) {
// variable is undefined or null
}
What you are checking this way is if foo is not:
false
Undefined
Null
+0, −0, or NaN
Empty String
More info here:
https://javascriptweblog.wordpress.com/2011/02/07/truth-equality-and-javascript/
This way your function is only executed if foo is valid.
You may want to be more specific and check against a certain type with typeof.
var foo = 'foo';
if (typeof foo === 'string') { // true
doSomething();
}
The typeof operator returns a string indicating the type of the
unevaluated operand.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/typeof
Let's say I have a variable myvar, and I don't have a variable myvar2. I can run the following without a problem:
typeof myvar
// ⇒ 'string'
typeof myvar2
// ⇒ 'undefined'
typeof and delete are the only functions I know of which don't throw errors when given an undefined parameter like this. I looked at the language spec for typeof and to my uninitiated eyes it seems to use internal functions like IsUnresolvableReference.
Edit: I'd been working in a language that checks type with a synonymous function, and hadn't noticed typeof is actually an operator in JavaScript. I've removed parentheses from the code here but left the above as written.
When I create a function:
function myFunc(input_variable) {
return("hello");
}
... as expected this throws a ReferenceError when passed myvar2 as a parameter, unless I run var myvar2;.
If I wrap the return in a try/catch statement to handle the myvar2 not defined case, I still get the same error, as the variable seems to be checked for a resolvable reference upon entry into the function (upon runtime?) :
function myFunc(input_var) {
try {
return "hello";
} catch(error) {
if (error.name === 'ReferenceError'){
return "world";
}
}
}
I was wondering how I can make a function that accepts unresolved references. My general guess is that, if it's a standard behaviour of functions, then perhaps I could modify some prototype for this construction specifically...? I'm aware prototypes are for objects, I'm wondering if this level of control over function is possible somehow?
By way of context, I always find myself writing function(input_var) :
if (typeof input_var == 'undefined' || my_settings.input_var_is_optional === true)
var input_var = 'Sometimes variables are optional. This is my default value.';
return dealWith(input_var);
} else if (typeof input_var == 'string') {
return dealWith(input_var);
} else {
// Already checked that input_var isn't optional, so we have a problem
return false; // or throw a TypeError or something like that
}
but the verbosity of all that plain puts me off writing type checking into my code, making it less robust to use functions more freely, or to pass onto other developers.
I'd like to write a type handling function, e.g.
For a function myFunc(input_var), if the variable passed in as parameter input_var has been defined, check if it's a string, else set it as "default_value". If it wasn't defined, also set it as "default_value", else it's a valid string, so just use input_var as is.
...but it's sabotaged by the fact that I can't actually pass anything in that's undefined, effectively stopping me from isolating this complexity in a separate function to which I could just pass 2 parameters: input_var (the real deal, not just its name), and expected_type.
function typeTest(input_var, expected_type) {
var is_optional_value = (typeof expected_type != 'undefined'
&& expected_type === true);
var optional_str = is_optional_value ? "|(undefined)" : ''
var type_test_regex = RegExp('^(?!' + expected_type + optional_str + '$)');
var is_expected_type = type_test_regex.test(typeof(input_var));
}
For example, to check that an optional variable passed into a function was both defined, and was defined as a string,
var myvar = 'abc'
// myvar2 is never defined
// Mandatory type (expecting a string):
typeTest(myvar, 'string'); // true
// if (/^(?!string)$)/.test(typeof(myvar))
typeTest(myvar2, 'string'); // throws error
// Mandatory type (expecting a number):
typeTest(myvar, 'number'); // false
typeTest(myvar2, 'number'); // throws error
// Optional type ("expected is true"):
typeTest(myvar, true); // true
// if (/^(?!string|(undefined)$)/.test(typeof(myvar))
typeTest(myvar2, true); // throws error
I was wondering how I can make a function that accepts unresolved references.
You can't. When you access an undeclared variable, the ReferenceError occurs before the function even gets called. There's nothing you can do inside the function to recover from this, because it hasn't even been called.
typeof and delete are the only functions I know of which don't throw errors when given an undefined parameter like this.
typeof and delete are not functions. That's why.
For example, to check that an optional variable passed into a function was both defined, and was defined as a string.
There's nothing stopping you from doing this. There is a difference between:
variables with the value undefined
parameters that have not been passed a value
undeclared variables.
There is no problem in dealing with the first two:
function hasType(val, type) {
return typeof val === type;
}
function myFunc(param1, param2) {
console.log('param1: ', hasType(param1, 'string'));
console.log('param2: ', hasType(param2, 'string'));
}
myFunc('hello');
There is no need to check whether someone is trying to call your functions with undeclared variables. If they are, then the problem is with their code and they need to fix it. If they are taking advantage of optional parameters, that is a different matter, and you can handle for that scenario just fine.
as the variable seems to be checked for a resolvable reference upon entry into the function
It is checked before entry.
Given foo(bar), the logic for resolution is "Get foo, then get bar, then call foo with the value of bar as an argument.
If bar isn't declared then you'll get a ReferenceError before the function is called in the first place.
typeof and delete are the only functions I know of which don't throw errors when given an undefined parameter like this.
From the documentation you link to:
The typeof Operator
The delete Operator
They aren't functions.
I was wondering how I can make a function that accepts unresolved references.
You can't.
For example, to check that an optional variable
If you want an argument to be optional then either:
Explicitly pass undefined:
typeTest(undefined, 'string');
Put the optional argument last in the arguments list:
typeTest('string');
Pass an object:
typeTest({ argument_name: 'string' });
You can using a function slide.
function attempt(f){
console.log(f());
}
attempt( function (){ return nomansland} );
//later an ajax call declares it:
var nomansland = "ok";
attempt( function (){ return nomansland} );
In the below code how does passing bar as
function (n) { return n; }
to foo evaluate to true in the if block?
function foo(bar) {
if (bar) {
// true
} else {
// false
}
}
This has me puzzled so any help is much appreciated.
If bar is bound to an anonymous function, then it is an object. Objects are 'truthy' in JavaScript.
The only values in JavaScript that are 'falsy' are:
false
null
undefined
'' (empty string)
0 (zero as a number)
NaN
Everything else is 'truthy', including function objects.
If you meant to call the anonymous function, you'd do if (bar(5)) which would call your anonymous function with the argument 5. Then your anonymous function would return n (which is 5 in this case). As 5 is not a falsy object, this would go to the true branch as well. Doing if (bar(0)) would got to the else branch, because 0 is falsy.
Anything not null, 0, false, empty string or undefined is going to evaluate to true in an if(something) statement, this is just how weak-typing in JavaScript works.
If you want more specificity you may want to look at the typeof operator to check for the type you're expecting, or use a another stronger check like this:
if(bar === true) {
Using === checks for both value and type equivalence.
You could use typeof() if you want to know what type it returns
It always return true, because bar is not null. if an expression within an if statement is not a logic expression (e.g. if(x <7)) then performs a check for null. If it is null it returns false, otherwise true.
In your example you have defined bar as the function {returns n;} so that is why your if statement is evaluating to true.
If bar returns a bool (true or false) then you need to call the function and get the result, rather than passing a reference to the function - this is done using parentheses:
var exampleA = bar(false); // executes function bar and returns the result (false) in exampleA
var exampleB = bar; // stores a reference to the function bar in variable exampleB
I have now seen 2 methods for determining if an argument has been passed to a JavaScript function. I'm wondering if one method is better than the other or if one is just bad to use?
function Test(argument1, argument2) {
if (Test.arguments.length == 1) argument2 = 'blah';
alert(argument2);
}
Test('test');
Or
function Test(argument1, argument2) {
argument2 = argument2 || 'blah';
alert(argument2);
}
Test('test');
As far as I can tell, they both result in the same thing, but I've only used the first one before in production.
Another Option as mentioned by Tom:
function Test(argument1, argument2) {
if(argument2 === null) {
argument2 = 'blah';
}
alert(argument2);
}
As per Juan's comment, it would be better to change Tom's suggestion to:
function Test(argument1, argument2) {
if(argument2 === undefined) {
argument2 = 'blah';
}
alert(argument2);
}
There are several different ways to check if an argument was passed to a function. In addition to the two you mentioned in your (original) question - checking arguments.length or using the || operator to provide default values - one can also explicitly check the arguments for undefined via argument2 === undefined or typeof argument2 === 'undefined' if one is paranoid (see comments).
Using the || operator has become standard practice - all the cool kids do it - but be careful: The default value will be triggered if the argument evaluates to false, which means it might actually be undefined, null, false, 0, '' (or anything else for which Boolean(...) returns false).
So the question is when to use which check, as they all yield slightly different results.
Checking arguments.length exhibits the 'most correct' behaviour, but it might not be feasible if there's more than one optional argument.
The test for undefined is next 'best' - it only 'fails' if the function is explicitly called with an undefined value, which in all likelyhood should be treated the same way as omitting the argument.
The use of the || operator might trigger usage of the default value even if a valid argument is provided. On the other hand, its behaviour might actually be desired.
To summarize: Only use it if you know what you're doing!
In my opinion, using || is also the way to go if there's more than one optional argument and one doesn't want to pass an object literal as a workaround for named parameters.
Another nice way to provide default values using arguments.length is possible by falling through the labels of a switch statement:
function test(requiredArg, optionalArg1, optionalArg2, optionalArg3) {
switch(arguments.length) {
case 1: optionalArg1 = 'default1';
case 2: optionalArg2 = 'default2';
case 3: optionalArg3 = 'default3';
case 4: break;
default: throw new Error('illegal argument count')
}
// do stuff
}
This has the downside that the programmer's intention is not (visually) obvious and uses 'magic numbers'; it is therefore possibly error prone.
If you are using jQuery, one option that is nice (especially for complicated situations) is to use jQuery's extend method.
function foo(options) {
default_options = {
timeout : 1000,
callback : function(){},
some_number : 50,
some_text : "hello world"
};
options = $.extend({}, default_options, options);
}
If you call the function then like this:
foo({timeout : 500});
The options variable would then be:
{
timeout : 500,
callback : function(){},
some_number : 50,
some_text : "hello world"
};
This is one of the few cases where I find the test:
if(! argument2) {
}
works quite nicely and carries the correct implication syntactically.
(With the simultaneous restriction that I wouldn't allow a legitimate null value for argument2 which has some other meaning; but that would be really confusing.)
EDIT:
This is a really good example of a stylistic difference between loosely-typed and strongly-typed languages; and a stylistic option that javascript affords in spades.
My personal preference (with no criticism meant for other preferences) is minimalism. The less the code has to say, as long as I'm consistent and concise, the less someone else has to comprehend to correctly infer my meaning.
One implication of that preference is that I don't want to - don't find it useful to - pile up a bunch of type-dependency tests. Instead, I try to make the code mean what it looks like it means; and test only for what I really will need to test for.
One of the aggravations I find in some other peoples' code is needing to figure out whether or not they expect, in the larger context, to actually run into the cases they are testing for. Or if they are trying to test for everything possible, on the chance that they don't anticipate the context completely enough. Which means I end up needing to track them down exhaustively in both directions before I can confidently refactor or modify anything. I figure that there's a good chance they might have put those various tests in place because they foresaw circumstances where they would be needed (and which usually aren't apparent to me).
(I consider that a serious downside in the way these folks use dynamic languages. Too often people don't want to give up all the static tests, and end up faking it.)
I've seen this most glaringly in comparing comprehensive ActionScript 3 code with elegant javascript code. The AS3 can be 3 or 4 times the bulk of the js, and the reliability I suspect is at least no better, just because of the number (3-4X) of coding decisions that were made.
As you say, Shog9, YMMV. :D
In ES6 (ES2015) you can use Default parameters
function Test(arg1 = 'Hello', arg2 = 'World!'){
alert(arg1 + ' ' +arg2);
}
Test('Hello', 'World!'); // Hello World!
Test('Hello'); // Hello World!
Test(); // Hello World!
url = url === undefined ? location.href : url;
There are significant differences. Let's set up some test cases:
var unused; // value will be undefined
Test("test1", "some value");
Test("test2");
Test("test3", unused);
Test("test4", null);
Test("test5", 0);
Test("test6", "");
With the first method you describe, only the second test will use the default value. The second method will default all but the first (as JS will convert undefined, null, 0, and "" into the boolean false. And if you were to use Tom's method, only the fourth test will use the default!
Which method you choose really depends on your intended behavior. If values other than undefined are allowable for argument2, then you'll probably want some variation on the first; if a non-zero, non-null, non-empty value is desired, then the second method is ideal - indeed, it is often used to quickly eliminate such a wide range of values from consideration.
I'm sorry, I still yet cant comment, so to answer Tom's answer...
In javascript (undefined != null) == false
In fact that function wont work with "null", you should use "undefined"
There is a tricky way as well to find, whether a parameter is passed to a function or not. Have a look at the below example:
this.setCurrent = function(value) {
this.current = value || 0;
};
This necessary means that if the value of value is not present/passed - set it to 0.
Pretty cool huh!
Why not using the !! operator? This operator, placed before the variable, turn it to a boolean (if I've understood well), so !!undefined and !!null (and even !!NaN, which can be quite interesting) will return false.
Here is an exemple:
function foo(bar){
console.log(!!bar);
}
foo("hey") //=> will log true
foo() //=> will log false
Sometimes you want undefined as a possible argument but you still have situations where the argument may not be passed. In that case you can use arguments.length to check how many arguments were passed.
// Throw error if the field is not matching our expectations
function testField(label, fieldValue, expectedValue) {
console.log(arguments) // Gives: [Arguments] { '0': 'id', '1': 1, '2': undefined }
if(arguments.length === 2) {
if(!fieldValue) {
throw new Error(`Field "${label}" must have a value`)
}
}
else if(expectedValue === undefined) {
if(fieldValue !== undefined) {
throw Error(`Field "${label}" must NOT have a value`)
}
}
// We stringify so our check works for objects as well
else {
if(JSON.stringify(fieldValue) !== JSON.stringify(expectedValue)) {
throw Error(`Field "${label}" must equal ${expectedValue} but was ${fieldValue}`)
}
}
}
testField('id', 12) -> Passes, we don't want id to be blank
testField('id', undefined, undefined) -> Passes, we want id to be undefined
testField('id', 12, undefined) -> Errors, we wanted id to be undefined
It can be convenient to approach argument detection by evoking your function with an Object of optional properties:
function foo(options) {
var config = { // defaults
list: 'string value',
of: [a, b, c],
optional: {x: y},
objects: function(param){
// do stuff here
}
};
if(options !== undefined){
for (i in config) {
if (config.hasOwnProperty(i)){
if (options[i] !== undefined) { config[i] = options[i]; }
}
}
}
}
Some times you may also want to check for type, specially if you are using the function as getter and setter. The following code is ES6 (will not run in EcmaScript 5 or older):
class PrivateTest {
constructor(aNumber) {
let _aNumber = aNumber;
//Privileged setter/getter with access to private _number:
this.aNumber = function(value) {
if (value !== undefined && (typeof value === typeof _aNumber)) {
_aNumber = value;
}
else {
return _aNumber;
}
}
}
}
function example(arg) {
var argumentID = '0'; //1,2,3,4...whatever
if (argumentID in arguments === false) {
console.log(`the argument with id ${argumentID} was not passed to the function`);
}
}
Because arrays inherit from Object.prototype. Consider ⇑ to make the world better.
fnCalledFunction(Param1,Param2, window.YourOptionalParameter)
If above function is called from many places and you are sure first 2 parameters are passed from every where but not sure about 3rd parameter then you can use window.
window.param3 will handle if it is not defined from the caller method.