splat

What is the correct name for operator * , as in function(*args) ? unpack, unzip, something else? In Ruby and Perl 6 this has been called "splat", and I think most people from those communities will figure out what you mean if you call it that. The Python tutorial uses the phrase "unpacking argument lists", which is long and descriptive. I haven't heard any other particular name for it in Python. I call it "positional expansion", as opposed to ** which I call "keyword expansion". Danilo Piazzalunga The Python Tutorial simply calls it 'the * -operator'. It performs unpacking of arbitrary

Consider, for example, squares = *map((2).__rpow__, range(5)), squares # (0, 1, 4, 9, 16) *squares, = map((2).__rpow__, range(5)) squares # [0, 1, 4, 9, 16] So, all else being equal we get a list when splatting on the lhs and a tuple when splatting on the rhs. Why? Is this by design, and if yes, what's the rationale? Or, if not, are there any technical reasons? Or is this just how it is, no particular reason? The fact that you get a tuple on the RHS has nothing to do with the splat. The splat just unpacks your map iterator. What you unpack it into is decided by the fact that you've used tuple

How do I splat across objects without using ECMA6 features ? Attempt function can(arg0, arg1) { return arg0 + arg1; } function foo(bar, haz) { this.bar = bar; this.haz = haz; } myArgs = [1,2]; With can I can just do: can.apply(this, myArgs); When trying with foo : new foo.apply(this, myArgs); I get this error (because I'm calling new ): TypeError: function apply() { [native code] } is not a constructor Using Object.create function foo(bar, haz) { this.bar = bar; this.haz = haz; } x = Object.create(foo.prototype); myArgs = [5,6]; foo.apply(x, myArgs); console.log(x.bar); Ven Using Object.create

This question already has an answer here: Optional argument after splat argument 6 answers I am able to define a method like this: def test(id, *ary, hash_params) # Do stuff here end But this makes the hash_params argument mandatory. These don't work either: def t(id, *ary, hash_params=nil) # SyntaxError: unexpected '=', expecting ')' def t(id, *ary, hash_params={}) # SyntaxError: unexpected '=', expecting ')' Is there a way to make it optional? You can't do that. You have to think about how Ruby would be able to determine what belongs to *ary and what belongs to the optional hash. Since Ruby

Run the following code, a = [1, 2, 3, 4, 5] head, *tail = a p head p tail You will get the result 1 [2, 3, 4, 5] Who can help me to explain the statement head,*tail = a , Thanks! head, *tail = a means to assign the first element of the array a to head , and assign the rest of the elements to tail . * , sometimes called the "splat operator," does a number of things with arrays. When it's on the left side of an assignment operator ( = ), as in your example, it just means "take everything left over." If you omitted the splat in that code, it would do this instead: head, tail = [1, 2, 3, 4, 5] p