Question

我有一个函数private: int numbers; int coffee;如下：

currentPlayer

我正在尝试在reducer中使用此函数，如下所示：

function currentPlayer(players) {
  return players.findIndex(player => player.current) || 0;
}

运行此代码时，出现以下错误：

ReferenceError：未定义currentPlayer

如果我按如下方式更改变量名称，则代码有效：

export function players(players = [], action) {
  players = players.slice();
  switch (action.type) {
    ...
    case DELETE_PLAYER:
      const currentPlayer = currentPlayer(players);
      if (currentPlayer === action.index) {
        players = switchNextPlayer(players, currentPlayer);
      }
      players.splice(action.index, 1);

  }
  return players;
}

我认为变量const currentPlayerIndex = currentPlayer(players);将在reducer函数中作用域。这是怎么回事？

Answer 1

在初始化表达式之前，const声明被解释为。因此，您的本地范围首先获得其自己的currentPlayer符号，然后评估初始化表达式。因此，在初始化表达式中，currentPlayer引用新声明的本地符号，而不是函数。

函数中未定义本地范围的变量

1 个答案: