Two javascript functions not working as same way

Question

I have written 2 javascript function but they are not working as same.

console.log(func2()); is undefined. Can anyone tell me why and how to solve this?

function func1()
{
  return {
      bar: "hello"
  };
}

function func2()
{
  return
  {
      bar: "hello"
  };
}

console.log(func1());
console.log(func2());

@T.J.Crowder: You have a list? Is it public? :) There's times I just know there has to be a dupe and yet I can't find a good one. — Amadan, Dec 16 '16 at 08:46
@Amadan: I do have an informal one, I just keep it in bookmarks. :-) — T.J. Crowder, Dec 16 '16 at 08:48

score 7 · Accepted Answer · answered Dec 16 '16 at 08:36

It's because of automatic semicolon insertion. Never put a newline after return and prior to what you want to return, it's treated as though it terminates the return statement (e.g., a ; is inserted after return), and your function ends up effectively returning undefined.

score 0 · Answer 2 · answered Dec 16 '16 at 08:37

0

I know this one. It's semicolon insertion. func2 is translated to

function func2()
{
  return;
  {
      bar: "hello"
  };
}

And return undefine.

answered Dec 16 '16 at 08:37

vothaison

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Two javascript functions not working as same way

2 Answers2