Checking for null or undefined VS handling errors VS returning bogus in JS

Question

similar question in Java: Java: check for null or allow exception handling

related:How to deep check for "null" or "undefined" with JS?

assuming a.foo() is a well-tested function that returns either

undefined or an object with a function such as {bar:()=>log('baz')},

hence, it would seem logical to write

a.foo().bar()

which obviously won't work.

(the undefined is the result of an empty return; when no information can be delivered)

Should

1. The operation be tried,

try{a.foo().bar();}catch(e){}

2. a new variable be declared

const u = a.foo();
if(u) u.bar();

3. or foo() return {bar:()=>{}} instead of undefined?

so that

a.foo().bar();

can be written normally?

what is the industry standard? what is fastest?

It is a personal opinion on what to do, there is no right answer. — epascarello
– epascarello, Commented Sep 23, 2017 at 2:12
There is no standard. I can tell you no one would use try catch — epascarello
– epascarello, Commented Sep 23, 2017 at 2:28

Kristianmitk · Accepted Answer · 2017-09-23 02:39:03Z

0

As already commented, this is a personal opinion and I would also stick to it that most people would try to avoid the try-catch story...

But anyway I wanted to add that you can also simply check:

a.foo() && a.foo().bar()

in case a.foo() will return undefined, then the right side wont be executed and thus this expression return undefined.

On the other side, if a.foo() is not undefined than the right side will be executed and this expression will evaluate to the result of a.foo().bar()

answered Sep 23, 2017 at 2:39

Kristianmitk

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2 Comments

epascarello Over a year ago

but that is bad because you are calling the method twice. If it is expensive it is not wise. If it is just a static property, that is a way to do it.

Kristianmitk Over a year ago

in that case, just assign a.foo() to a variable and substitute a.foo() with the variable name. Than you call it just once as in a try-catch

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