我试图了解如何使用break
语句执行do / while循环。我以为只要执行break语句(前两次迭代,因为那时候if条件为true,但第三次都不是),for循环就会逸出,因此n不会递增。但是行console.log(`n is ${n}`)
记录了n is 2
n is 2
n is 5
-它如何从2跳到5?我以为第三次是3(如果if条件不成立,因为quot
为60
,而60 % 2
等于0,因此break
语句未执行)。
// newArray = [5, 4, 3, 2, 1];
var quot = 0;
var loop = 1;
var n;
do {
quot = newArr[0] * loop * newArr[1];
for (n = 2; n < newArr.length; n++) {
if (quot % newArr[n] !== 0) {
break;
}
}
console.log(`n is ${n}`)
loop++;
} while (n !== newArr.length);
这是完整的代码(freeCodeCamp challenge的解决方案):
function smallestCommons(arr) {
// Sort array from greater to lowest
// This line of code was from Adam Doyle (http://github.com/Adoyle2014)
arr.sort(function(a, b) {
return b - a;
});
// Create new array and add all values from greater to smaller from the
// original array.
var newArr = [];
for (var i = arr[0]; i >= arr[1]; i--) {
newArr.push(i);
}
// Variables needed declared outside the loops.
var quot = 0;
var loop = 1;
var n;
// Run code while n is not the same as the array length.
do {
quot = newArr[0] * loop * newArr[1];
for (n = 2; n < newArr.length; n++) {
if (quot % newArr[n] !== 0) {
break;
}
}
console.log(`n is ${n}`)
loop++;
} while (n !== newArr.length);
return quot;
}
// test here
smallestCommons([1,5]);
答案 0 :(得分:0)
您的数组长度为5,因此,如果未执行if语句和break,则n始终为5。
(n = 2; n 上一次执行此操作, 就像您在问题中说的那样:“ if条件不成立,因为quot为60,而60%2等于0,因此不会执行break语句” 您的break语句只会使您早日退出for循环,如果您的for循环完成了整个周期,则n始终为5。 希望如此!n=4, n < newArr.length (which is 5), n++ (now n is 5)