Question

I don’t understand why fooA and fooB outcome different.

var foo = function(){}

fooA = new foo();

foo.prototype.x = 1;
foo.prototype = { y: 2, z: 3};

console.log(fooA.x, fooA.y, fooA.z);// 1, undefined, undefined

fooB = new foo();
console.log(fooB.x, fooB.y, fooB.z);// undefined, 2, 3

1. does foo.prototyp = {} override the method defined in front of it?

2. Why fooA is state in front of prototype.x, it inherit the result, but not y and z?

Answer 1

The reason for that behavior is that you make foo.prototype point to a new object when using foo.prototype = { y: 2, z: 3};, and existing objects’ prototypes don’t change when the constructor’s prototype property is set to a new value.

A line-by-line explanation of what happens:

var foo = function(){}

foo.prototype is initialized as an empty object (we shall call this object A).

fooA = new foo()

fooA is set to a new foo object, it has its prototype set to foo.prototype (A).

foo.prototype.x = 1

Because fooA‘s prototype is the same object as foo.prototype, fooA.x becomes 1. In other words, A gets the property x = 1.

foo.prototype = { y: 2, z: 3};

We create a new object that has the properties y = 2 and z = 3. We shall call this object B. foo.prototype is set to the new object.

console.log(fooA.x, fooA.y, fooA.z);

fooA‘s prototype is still A, which only has the x = 1 property.

fooB = new foo();

We create a new foo object whose prototype is B.

console.log(fooB.x, fooB.y, fooB.z);

fooB‘s prototype is B, which has the properties y = 2 and z = 3, but not the property x.