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  • [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
    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]
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  • 原文地址:https://www.cnblogs.com/Answer1215/p/11688327.html
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