Why doesn't a Javascript return statement work when the return value is on a new line?

Consider the following JavaScript:

function correct()
{
    return 15;
}

function wrong()
{
    return
          15;
}

console.log("correct() called : "+correct());
console.log("wrong() called : "+wrong());

The correct() method in the above code snippet returns the correct value which is 15 in this case. The wrong() method, however returns undefined. Such is not the case with the most other languages.

The following function is however correct and returns the correct value.

function wrong()
{
    return(
          15);
}

If the syntax is wrong, it should issue some compiler error but it doesn't. Why does this happen?

Answers:

Answer

Technically, semi colons in javascript are optional. But in reality it just inserts them for you at certain newline characters if it thinks they are missing. But the descisions it makes for you are not always what you actually want.

And a return statement followed by a new line tells the JS intepreter that a semi colon should be inserted after that return. Therefore your actual code is this:

function wrong()
{
    return;
          15;
}

Which is obviously wrong. So why does this work?

function wrong()
{
     return(
           15);
}

Well here we start an expression with an open(. JS knows we are in the middle of an expression when it finds the new line and is smart enough to not insert any semi colons in this case.

Answer

If there is nothing after the return statement on that line then ; will be inserted there which will result in returning without any values => return value is undefined.

See: http://lucumr.pocoo.org/2011/2/6/automatic-semicolon-insertion/

Answer

The command line of the javascript can not be broken by line breaks. But arguments of functions can be broken, not highly recommended (done in your example).

Tags

Recent Questions

Top Questions

Home Tags Terms of Service Privacy Policy DMCA Contact Us

©2020 All rights reserved.