问题描述
这里我必须缺少一些基本原理，但是函数shuffle可以按预期工作，但是当我尝试在另一个函数中多次调用任何函数时，第一次调用会导致数组中充满52个未定义的条目。
function shuffle(array) {
let newDeck = [];
for (i=52; i>0; i--) {
let randomPick = Math.floor((Math.random() * array.length));
newDeck.push(array[randomPick]);
deck.splice(randomPick,1);
}
deck = newDeck;
console.log(deck);
}
function fullShuffle(cards) {
shuffle(cards);
shuffle(cards);
}
解决方法
该代码在问题中并不完整，但似乎deck是一个全局变量，您还可以将其作为参数传递给fullShuffle。问题是deck.splice在一次调用shuffle中就完全清空了该数组，因此，如果再次使用该deck并将其通过fullShuffle传递给{{1}再一次，}，您传递的是一个空数组，并且所有52次迭代都未定义shuffle（因为如果我对您其余部分的假设正确，则array[randomPick]与array的引用相同）
如果您想通过这种方式实施改组，则：
- 请勿在该函数内使用全局变量
- 仅依靠函数获取的参数
- 获取该数组的副本，以便您可以愉快地拼接而不影响调用方的数组
- 将经过改组的数组返回给调用方
代码：
deck尽管对于人类洗牌来说，将同一副牌洗牌两次是有意义的，但是对于这样的功能，两次调用它几乎没有意义。就像您不信任随机随机播放的功能一样。
还要注意，function shuffle(array) {
let newDeck = [];
array = [...array]; // take copy!
for (i=52; i>0; i--) {
let randomPick = Math.floor((Math.random() * array.length));
newDeck.push(array[randomPick]);
array.splice(randomPick,1);
}
console.log(newDeck);
return newDeck; // return it.
}
function fullShuffle(cards) {
cards = shuffle(cards); // capture the returned,shuffled array
cards = shuffle(cards);
return cards; // return to caller
}
// can be called like this:
var deck = [ "ac","2c","3c","4c","5c","6c","7c","8c","9c","10c","jc","qc","kc","ad","2d","3d","4d","5d","6d","7d","8d","9d","10d","jd","qd","kd","ah","2h","3h","4h","5h","6h","7h","8h","9h","10h","jh","qh","kh","as","2s","3s","4s","5s","6s","7s","8s","9s","10s","js","qs","ks" ];
deck = fullShuffle(deck);
有更好的实现。看看here，您会找到Durstenfeld算法的实现。