简约失态

Question

当我写这篇文章时，我假设第二个控制台日志将是一个4的数组。我的印象是，我的代码将带谁，然后将其连接到数组2的第一个console.log，然后下一个迭代将是一个4的数组，因为它是连接的。

let player1Hand = [];



function drawDeck(whom, whichDeck, howMany) {
    whom = whom.concat(whichDeck.splice(0, howMany));
    console.log(whom);
}

drawDeck(player1Hand, genericDeck, 2);
drawDeck(player1Hand, genericDeck, 2);

相反，控制台日志是第一个数组，然后第二个日志是一个分隔数组。我想当我写这篇文章的时候，我自己躲起来了。

Answer 1

您遇到的问题是使用concat返回一个新数组，而不是将这些项添加到您已经拥有的当前数组中。将其设置为旧值不会更新引用。所以你应该和推荐人合作。

使用ES6，您可以使用spread运算符

const newItems = whichDeck.splice(0, howMany)
whom.push(...newItems);

或者没有ES6，你可以用

var newItems = whichDeck.splice(0, howMany)
whom.push.apply(whom, newItems)

​

let playerDeck1 = []
let playerDeck2 = []
let remainingCards = [1,2,3,4,5,6,7,8,9,
10,11,12,13,14,15,16,17,18,19,
20,21,22,23,24,25,26,27,28,29]

const deal = (playerDeck, pile, drawCount) => {
  const newItems = pile.splice(0, drawCount)  // remove top N cards from the deck
  playerDeck.push(...newItems);  // put them on the end of the players current hand
}

console.group("turn1")
deal(playerDeck1, remainingCards, 5)  // 1,2,3,4,5
deal(playerDeck2, remainingCards, 5)  // 6,7,8,9,10
console.log("player1", JSON.stringify(playerDeck1))
console.log("player2", JSON.stringify(playerDeck2))
console.groupEnd("turn1")


console.group("turn2")
deal(playerDeck1, remainingCards, 2) // 1,2,3,4,5 + 11,12
deal(playerDeck2, remainingCards, 2) // 6,7,8,9,10 + 13,14
console.log("player1", JSON.stringify(playerDeck1))
console.log("player2", JSON.stringify(playerDeck2))
console.groupEnd("turn2")

console.group("turn3")
deal(playerDeck1, remainingCards, 2) // 1,2,3,4,5,11,12 + 15,16
deal(playerDeck2, remainingCards, 2) // 6,7,8,9,10,13,14 + 17,18
console.log("player1", JSON.stringify(playerDeck1))
console.log("player2", JSON.stringify(playerDeck2))
console.groupEnd("turn3")

​

Answer 2

let player1Hand = [];



function drawDeck(whom, whichDeck, howMany) {
    whom = whom.concat(whichDeck.splice(0, howMany));
    console.log(whom);
    player1Hand = whom;
}

drawDeck(player1Hand, genericDeck, 2);
drawDeck(player1Hand, genericDeck, 2);

我只是躲着自己..。在这个函数是临时的，我需要重新评估player1Hand的值，让它回到函数的范围之外。因此，第二次运行时，player1Hand值已更改为第一项.

抱歉的。也许这会对其他人有所帮助。

Answer 3

  whom = whom.concat(whichDeck.splice(0, howMany));

它用一个新数组重写存储到局部变量whom中的引用。它不以任何方式更改传递的引用或外部变量。因此，playerHand1不会改变。相反，您可以扩展到Array.push中来实际地对传递的数组进行变异：

 whom.push(...whichDeck.splice(0, howMany));

或者只返回新的数组：

return whom.concat(whichDeck.splice(0, howMany));

然后将其存储在playerHand1中

  playerHand1 = drawDeck(playerHand1, genericDeck, 2);