Why can't println(); be used as a variable?

Question

I'm fairly new to Javascript, and am confused on something. Why can't the command "println("..."); be called as a variable such as: var num = println("...");. I could be wrong, and if you are able to, I'd be happy to know how. But after some testing it seems like I can't. My test code is:

function start() {
    var SENTINEL = "1 1";
    var rollOne = Randomizer.nextInt(1, 6);
    var rollTwo = Randomizer.nextInt(1, 6);
    var num = println(rollOne + rollTwo);
    if(num == SENTINEL) {
          println("You did it");
    }
}

All it's supposed to do is give to random numbers in a # # form and, if it sees that the numbers are 1,1, it will give a message. It wont give the message and can't seem to view the variable "num" as an actual variable. But when I change the variable num to simply asking the user for a number:

function start() {
    var SENTINEL = -1;
    var rollOne = Randomizer.nextInt(1, 6);
    var rollTwo = Randomizer.nextInt(1, 6);
    var num = readInt("Enter number");
    if(num == SENTINEL) {
        println("You did it");
}

} And type in -1, it triggers the sentinel, thus promptly displaying the message. This is a really roundabout way to ask a simple question but I hope I can get some help. Thank you :)

Answer 1

Why can't the command "println("..."); be called as a variable such as: var num = println("..."); [...] It wont give the message and can't seem to view the variable

If the value returned is unusable, it is most likely undefined; i.e. The function println doesn't explicitly return anything.

In your case, you could try something like this:

var printInt = function(num) { println(num); return num; }

Note, println isn't part of the standard JavaScript language. For modern web browsers, it can be adapted to use (console.log(...)).

var printInt = function(num) { console.log(num); return num; }

And then to adapt to your code:

var num = printInt(rollOne + rollTwo);

But this still won't validate because you're comparing against "1 1" whereas your logic will return 2. JavaScript (as well as many other languages) implicitly uses addition when supplied with two numbers, but concatenation when supplied with at least one string.

var SENTINEL = "1 1";  // <---- String!
var SENTINEL = -1;     // <---- Number!

So you should consider something like this instead (renamed accordingly):

var printRolls = function(text) { println(text); return text; }
var rolls = printRolls(rollOne + " " + rollTwo);
if(rolls == SENTINEL) {
      println("You did it");
}

Or to simplify it a bit:

if(printRolls(rollOne + " " + rollTwo) == SENTINEL)
    println("You did it");

Answer 2

It is possible that println doesn't return the string that is passed into. In that case, you can use

if (SENTINEL === rollOne + " " + rollTwo)

to format the string and properly test equality.

Answer 3

In JavaScript it is possible to assign the return value from any function to a variable similar to how you've done it:

var anyVariable = anyFunction();

But, some functions return the value undefined. Or they return a number, or an array, or...whatever.

I imagine your println() function prints the value you pass to it somewhere (on the screen? to the console?) and then returns undefined. Or if it is returning the printed value it is in a format different to what you have used in your SENTINEL variable. So then when you try to compare that with SENTINEL it won't be equal.

To fix your original function, assign the sum of the rolls to a variable, then print and test that:

function start() {
    var SENTINEL = 2;
    var rollOne = Randomizer.nextInt(1, 6);
    var rollTwo = Randomizer.nextInt(1, 6);
    var num = rollOne + rollTwo;
    println(num);
    if(num == SENTINEL) {
          println("You did it");
    }
}

EDIT: if you want the println() to display a string like "1 1" or "3 5" to show what each of the two rolls were then do this:

println(rollOne + " " + rollTwo);

That is, create a new string that is the result of concatenating rollOne's value with a single space and then rollTwo's value.