Unusual javascript syntax

Question

could someone please explain to me what is going on in the second line here ? :

var foo = function(){alert("hello?")}; (0,foo)(); 

Answer 1

The infamous comma expression a,b evaluates both arguments and returns the value of the right-hand expression.

Hence in this case it's exactly the same as foo();.

Here's a better example that will help you understand what's going on:

function foo() {     print("foo called");     return 123; } function bar() {     print("bar called");     return 456; } var result = (foo(), bar()); print("result:", result); 

Output:

foo called bar called result: 456 

Also the comma expression may be confused with the comma delimiting function arguments. Not the same! Notice the difference:

print("result:", foo(), bar() ); // 3 arguments, no comma operator print("result:", (foo(), bar()) ); // 2 arguments, comma operator 

Answer 2

It's evaluating both expressions in the first parenthesis and executing the second one (in this case - a function).

https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Operators/Comma_Operator