Why would a function-call expression use a comma-operator expression as the function to be called?

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

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

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