当我将changeInValue设置为0时，缓动函数始终返回初始值

Question

// t: current time, b: begInnIng value, c: change In value, d: duration
easeOutSine: (t, b, c, d) => c * Math.sin(t/d * (Math.PI/2)) + b;

Obviously, if the parameter of 'c' is 0, that will always return the begining value, so what should I do if I just want change the value to 0?

Answer 1

调用函数c （值的变化）等于-b （起始值的负数）。