[Javascript] Avoid Accidental Returns of New State by using the void Keyword

For example we have a 'useState' function, which takes a state and a function to update the state:

const useState = (state, setState) => {
  const newState = setState(state);
  if (newState != null) {
    return newState;
  } else {
    return state;
  }
};

If the new state is not undefined or null, we will return newState otherwise, we return the original state.

 

But when we run the code like this:

const res = useState([1], state => state.push(2)); // 2

We expect the res to be [1, 2], but we got 2, this is because 'push' method return the length of the array as a result.

 

To solve the problem we can use 'void' keyword, it will execute the expression and return undefined as a result, for example:

void 2 == '2' // (void 2) == '2', the same as undefined == '2', which is false
void (2 == '2') // void false which is undefined

 

复制代码
const useState = (state, setState) => {
  const newState = setState(state);
  if (newState != null) {
    return newState;
  } else {
    return state;
  }
};

const res = useState([1], state => void state.push(2));
console.log(res); //[1,2]
复制代码

 

 

posted @   Zhentiw  阅读(134)  评论(0编辑  收藏  举报
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