Question

当我写这篇文章的时候，我假设第二个控制台日志是一个4的数组。我的印象是我的代码会占用谁然后将其连接到第一个console.log的数组2然后下一次迭代将是一个4的数组，因为它是concat。

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单独的数组......我想我写这篇文章的时候，我自己橡皮擦了。

Answer 1

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);

我只是橡皮筋自己...在函数内部是临时的我需要重新评估player1H的值，然后让它指出函数的范围。所以第二次运行时，player1Hand值已经改为第一次尝试...
抱歉。也许这会在某些时候帮助别人。

Answer 2

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

用新数组重写存储在局部变量whom中的引用。它不会以任何方式更改传入的引用或外部变量。因此，playerHand1不会改变。相反，你可以传播到Array.push实际改变传递的数组：

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

或只返回新数组：

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

然后将其存储在playerHand1：

中

  playerHand1 = drawDeck(playerHand1, genericDeck, 2);

Answer 3

您遇到的问题是使用concat返回一个新数组，而不是将项添加到您已有的当前数组中。将其设置为旧值不会更新引用。所以你应该使用参考。

使用ES6，您可以使用点差运算符

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

或没有ES6，您可以使用

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")

Concat简单的misque

3 个答案: