Question

让我们说我想要＆＃34;数组中的6个元素仅包含3.如果到达结尾，则重新开始。

示例：

let arr = ["a", "b", "c"];
let wanted = 5;

for(let i = 0; i < wanted; i++) {
    console.log(arr[i]);
}

当然会返回：

a
b
c
undefined
undefined

相反，我需要：

a
b
c
a
b

有没有办法在元素结束后从数组的开头开始？

Answer 1

使用%(Remainder)运算符获取余数。

console.log(arr[i % arr.length]);

余数值将在数组索引范围内。

例如：当i到达

时

3 =＆gt; 3 % 3 = 0
4 =＆gt; 4 % 3 = 1
0 =＆gt; 0 % 3 = 0。

＆＃13;

let arr = ["a", "b", "c"];
let wanted = 5;

for (let i = 0; i < wanted; i++) {
  console.log(arr[i % arr.length]);
}

＆＃13;

Answer 2

您可以使用(%) remainder运算符。它将从0开始索引，直到arr.length - 1。

第一步是

[0 % 3] === 0 whole 0 remainder, 
[1 % 3] === 0 whole 1 remainder,
[2 % 3] === 0 whole 2 remainder
[3 % 3] === 1 whole 0 remainder
[4 % 3] === 1 whole 1 remainder

let arr = ["a", "b", "c"];
let wanted = 5;

for(let i = 0; i < wanted; i++) {
    console.log(arr[i % arr.length]);
}

Answer 3

只需使用%

即可

let arr = ["a", "b", "c"];
let wanted = 5;

for(let i = 0; i < wanted; i++) {
    console.log(arr[i % arr.length]);
}

Answer 4

let arr = ["a", "b", "c"];
let wanted = 5;    

for (let i = 0; i < wanted; i++) {
    console.log(arr[i]);
    if (arr.length == i) {
        i = 0;
    }
}

Answer 5

if (i+1 == arr.length){
 console.log(arr[i]);
 i=0;
 wanted =wanted-arr.length;
}else{
 console.log(arr[i]);
}

循环将正常工作，直到你的计数器位于最后一个元素，然后它将重置为0以重新开始，并将继续执行此操作，直到达到所需的元素数量。

当循环中没有更多元素时如何重新开始？

5 个答案: