Return a Function With Expanded Variables

Asked Jun 04 '16 at 00:22

Active Jun 04 '16 at 00:22

Viewed 24 times

This is a simplified example, but the goal is to have a function that returns a function given a set of arguments, and the returned function should have those arguments expanded into numbers/strings.

The example here returns a function that checks whether a set of coordinates are within a rectangle:

function createFunction(startX, startY, width, height) {
    return function(x, y) {
        if (x >= startX && x <= startX+width && y >= startY && y <= startY+height) {
            return true;
        }
        return false;
    };
}

var myFunction = createFunction(5, 10, 200, 100);

console.log(myFunction); //startX and the other arguments aren't expanded

The problem described in the comment above is that the arguments from the "createFunction" function aren't expanded into values, and if one used the returned function it would create an error because startX and the other arguments aren't globally defined.

How do I return a function with the variables expanded?

asked Jun 04 '16 at 00:22

Tom Burris

1

they don't need to be expanded, they are encapsulated. google `closures`. – Thomas Jun 04 '16 at 00:27
They are private variables, which are not known outside. – Praveen Kumar Purushothaman Jun 04 '16 at 00:27
Your code will run very fine. – gu mingfeng Jun 04 '16 at 00:27
What do you mean they are not expanded? You wont see them in the console log, but they are there and defined. For example, put a log statement in your `function (x, y){ ... }` and then call `myFunction(0, 1);` – Ryan Wheale Jun 04 '16 at 00:27

Return a Function With Expanded Variables

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