valueOf函数设置为始终返回相同的值

Question

This:

% node
> var o = {valueOf: function() { return 5; } };
undefined
> o += 1;
6
> o
6
> o += 1;
7
> o
7

Why is o value incremented if valueOf always returns 5?

Answer 1

Starting from the situation where o references the object as seen in the question:

var o = {valueOf: function() { return 5; } };

Then in the expression

o += 1;

what happens is:

1. The value of o is obtained as a number, via the call to .valueOf()
2. That value is added to 1 , giving 6
3. That result is assigned to the variable o

Thus the variable o , which once contained a reference to an object, now contains a number.

Answer 2

This is because you are assigning o = 5 , then you are incrementing the value, by doing += 1 and then you're querying the new value, which has now changed, increased by one, hence o = 6 . You're switching from object to a value when you're doing the first incrementation on the object!

Answer 3

The valueOf() method returns the primitive value of the specified object. What happened above is,

1. o is first assigned as object as below,

   var o = {valueOf: function() { return 5; } };

2. When you use o += 1; , you are here using o as primitive type and JavaScript calls the valueOf method to convert an object in to a primitive type

3. Hence, o is no longer object but converted as primitive type with value of 6 = 5 + 1 and started incremented with o += 1;

Hope it clarifies.