请问一下js ex6中 extends加constructor中的super和没有加有区别吗？

Question

        class Father {
            constructor(x, y) {
                this.x = x;
                this.y = y;
            }
            sum() {
                return this.x + this.y;
            }
        }
        class Son extends Father {
            subtract() {
                return this.x - this.y;
            }
        }
        var a = new Son(1, 1);
        console.log(a.sum());
        console.log(a.subtract());

这是没有加的控制台输出正常

        class Father {
            constructor(x, y) {
                this.x = x;
                this.y = y;
            }
            sum() {
                return this.x + this.y;
            }
        }
        class Son extends Father {
            constructor(x, y) {
                super(x, y);
            }
            subtract() {
                return this.x - this.y;
            }
        }
        var a = new Son(1, 1);
        console.log(a.sum());
        console.log(a.subtract());

这是加过了之后，控制台输出依然正常

请问这两种有区别吗？

Answer 1

constructor不写的话，子类也会把参数传给父类。如果写了constructor的话，super就一定要写了，不然参数无法传递给父类。