What does ReturnIfAbrupt mean in ES6 draft?

Question

I'm currently implementing some shims for the ES6 draft. I'm wondering if anyone can tell me what ReturnIfAbrupt means. For instance, my implementation for Number.toInt (which calls internal [[ToInteger]] is as follows:

if (!('toInt' in Number))
    Object.defineProperty(Number, 'toInt', {

        value: function toInt(value) {
            // ECMA-262 Ed. 6, 9-27-12. 9.1.4

            // 1. Let number be the result of calling ToNumber on the input argument.
            var number = Number(value);

            // 2. ReturnIfAbrupt(number).
            // ?

            // 3. If number is NaN, return +0.
            if (number != number) return 0;

            // 4. If number is +0, -0, +Infinity, or -Infinity, return number.
            if (number == 0 || 1 / number == 0) return number;

            // 5. Return the result of computing sign(number) * floor(abs(number)).
            return (n < 0 ? -1 : 1) * Math.floor(Math.abs(number));

        },

        writable: true,
        configurable: true

    });

Step 2 is ReturnIfAbrupt(number). You'll notice I currently have // ? for that step because I'm not sure what to do. What does it mean when it says ReturnIfAbrupt(...)?

I have read the section on ReturnIfAbrupt in the draft, however I am unable to understand what to do for step 2, what to put in place of // ? in the code above.

From my reading, it may be that nothing should be done, and the ReturnIfAbrupt step merely means to let any error which occurred in ToNumber to propagate up, exiting the function. However, that seems overly verbose, as I would think it could go without saying. Also, it doesn't seem to me like ToNumber can even throw an error. Could someone confirm or help me to understand the real meaning?