Javascript function fails to return object when there is a line-break between the return statement and the object?

Question

Here is the jsfiddle

The full code:

function foo1(){
    return {msg: "hello1"};
}
function foo2(){
    return
    {msg: "hello2"};
}

// output = "foo1 =  {"msg":"hello1"}"
console.log('foo1 = ' , JSON.stringify(foo1())); 

//output = " foo2 =  undefined "
console.log('foo2 = ' , JSON.stringify(foo2()));

The difference between the two is that in foo2, the {msg: 'hello'} is in its own new line. I was expecting the parser to ignore whitespaces?

Answer 1

tl;dr The line break is causing the 'undefined' in the second function. JavaScript doesn't require semicolons in a lot of cases and just assumes them in certain contexts (Automatic Semicolon Insertion).

---

In certain cases, to avoid this ASI problem and for aesthetic reasons, I use the so-called grouping operator. For example, with the hoquet templating DSL (it compiles Arrays as s-expressions into HTML), I like to group the Arrays so that they clearly show the HTML structure:

return (
  ["ul"
  , ["li"
    , ["span", {class: "name"}, this.name]
    , ["span", {id: "x"}, "x"]
    ]
  ]
);

which seems somewhat clearer, or more consistent to me than

return [
  "ul",
  [
    "li",
    ["span", {class: "name"}, this.name],
    ["span", {id: "x"}, "x"]
  ]
];

and they end up with the same number of lines. But it's a matter of aesthetics, really.

---

The grouping operator just evaluates whatever expression is inside of it. You see this commonly with Immediately Invoked Function Expressions, where you need to turn what would normally be a function declaration into an expression, which can then be immediately invoked (hence, the name). However, a perhaps lesser known feature of the grouping operator is that it can also take a comma-delimited list of expressions, e.g.

function() {
  return (
    doSideEffects(),
    console.log("this is the second side effect"),
    1 + 1
  );
}

In this case it evaluates each of these expressions and returns only the last one (1 + 1).