Javascript代码求和素数的循环异常

我有下面的代码，我试图将所有的素数相加到10

我不想为这个问题找到最有效的代码或正确的代码，但我很难理解for循环是如何工作的。我的具体意思是当

I=7

时

由于我有两个嵌套的for循环，出于某种原因，

I=7

外循环不止一次循环，我的意思是，内循环似乎达到了它的终止条件

j

，此时

j=2

和

k=2

，但它似乎坚持继续循环

代码如下：

var array = [2];
var total = 0; 
function isPrime(i, j) {
    if ( i%array[j] === 0 ) {
        console.log("Not P Check: i was " + i + ", j was " + j + " and k was " + k);
        console.log(i + " is not a prime");
        k = j;
    }
    else if ((j + 1) === array.length) {
        console.log(i + " is a prime");
        total += i;
        console.log("total so far is " + total);
        array.push(i);
        console.log(array);
        k = j;
        console.log("is P Check: i was " + i + ", j was " + j + " and k was " + k);
    }
    else {
        j++;
        console.log("Check " + (j + 1) + ": i is " + i + ", j is " + j + " and k is " + k);
        isPrime(i,j);
    }
}

for(var i = 3; i <=10; i++) {
   var k = array.length;
   for(var j = 0; j < k; j++) {
       console.log("Check 1: i is " + i + ", j is " + j + " and k is " + k);
       isPrime(i, j);
   }
}

console.log(total);
console.log(array);

var数组=[2]；
var合计=0；
函数isPrime（i，j）{
if（i%数组[j]==0）{
log（“非P检查：i为“+i+”，j为“+j+”，k为“+k”）；
log（i+“不是素数”）；
k=j；
}
else if（（j+1）==数组长度）{
log（i+“是素数”）；
总数+=i；
console.log（“到目前为止的总数为”+总数）；
阵列推送（i）；
console.log（数组）；
k=j；
log（“是P检查：我是“+i+”，j是“+j+”，k是“+k”）；
}
否则{
j++；
log（“检查”+（j+1）+”：i是“+i+”，j是“+j+”，k是“+k”）；
isPrime（i，j）；
}
}
对于（var i=3；i我为您创建了一个小提琴并编辑了几行。基本上，我更改了isPrime以返回一个布尔值，并告诉循环在找到素数后中断。在您的版本中，当它达到7时，7被推入数组（array.length增加）现在，当我们已经发现，7是一个素数，我们想打破我们的内部循环，继续使用不同的i

var array = [2];
var total = 0; 
function isPrime(i, j) {
    if ( i%array[j] === 0 ) {
        console.log("Not P Check: i was " + i + ", j was " + j + " and k was " + k);
        console.log(i + " is not a prime");
        k = j;
        return false;
    }
    else if ((j + 1) === array.length) {
        console.log(i + " is a prime");
        total += i;
        console.log("total so far is " + total);
        array.push(i);
        console.log(array);
        k = j;
        return true;
        console.log("is P Check: i was " + i + ", j was " + j + " and k was " + k);
    }
    else {
        j++;
        console.log("Check " + (j + 1) + ": i is " + i + ", j is " + j + " and k is " + k);
        return isPrime(i,j);
    }
}

for(var i = 3; i <=10; i++) {
   var k = array.length;
   for(var j = 0; j < k; j++) {
       console.log("Check 1: i is " + i + ", j is " + j + " and k is " + k);
       if (isPrime(i, j)) break;
   }
}

console.log(total);
console.log(array);

var数组=[2]；
var合计=0；
函数isPrime（i，j）{
if（i%数组[j]==0）{
log（“非P检查：i为“+i+”，j为“+j+”，k为“+k”）；
log（i+“不是素数”）；
k=j；
返回false；
}
else if（（j+1）==数组长度）{
log（i+“是素数”）；
总数+=i；
console.log（“到目前为止的总数为”+总数）；
阵列推送（i）；
console.log（数组）；
k=j；
返回true；
log（“是P检查：我是“+i+”，j是“+j+”，k是“+k”）；
}
否则{
j++；
log（“检查”+（j+1）+”：i是“+i+”，j是“+j+”，k是“+k”）；
返回i，j；
}
}
对于（var i=3；i来说，答案是在您的解决方案中不需要内环。您的递归函数已经在做内环的工作。内环中发生的情况是，一些数字的重复次数超过了需要的次数

Dygestor的解决方案是一种方法，另一种方法更简单

for(var i = 3; i <= 10; i++) {
  isPrime(i, 0);
}

删除内部循环的日志（请记住，内部循环中也缺少
控制台.log
）：

另外，在新的解决方案中，
k
变量是不必要的。
j
将递增，直到它到达素数数组的末尾或直到目标数可除为止，以较早者为准

如果要使用内部循环解决方案，则需要删除递归函数并执行以下操作：

var primes = [2];
var sum = 0;

// Start looping
for(var i = 3; i <= 10; ++i) {
  var prime = true; // Prime until proven innocent
  for(var j = 0; j < primes.length; ++j) { // Length stays same until later
    if(i % arr[j] === 0) { // The meat of your isPrime function: divisible?
      prime = false;
      break; // Stop the loop early: number is not prime!
    }
  }
  if(prime) { // We have a prime!
    primes.push(i); // Add it to our list of primes
    sum += i; // Add the prime to the sum
  }
}

// Log the result
console.log("The sum of primes up to 10 (inclusive)", sum);
console.log("These primes were", primes);

var primes=[2]；
var总和=0；
//开始循环
对于（var i=3；i）您是如何确定它不止一次循环的？看起来您的内部循环是不必要的，您的iPrime函数已经在运行了吗？@jack i刚刚在循环中添加了一个console.log行
3 is a prime
total so far is 3
[2, 3]
is P Check: i was 3, j was 0 and k was 0
Not P Check: i was 4, j was 0 and k was 0
4 is not a prime
Check 2: i is 5, j is 1 and k is 0
5 is a prime
total so far is 8
[2, 3, 5]
is P Check: i was 5, j was 1 and k was 1
Not P Check: i was 6, j was 0 and k was 1
6 is not a prime
Check 2: i is 7, j is 1 and k is 0
Check 3: i is 7, j is 2 and k is 0
7 is a prime
total so far is 15
[2, 3, 5, 7]
is P Check: i was 7, j was 2 and k was 2
Not P Check: i was 8, j was 0 and k was 2
8 is not a prime
Check 2: i is 9, j is 1 and k is 0
Not P Check: i was 9, j was 1 and k was 0
9 is not a prime
Not P Check: i was 10, j was 0 and k was 1
10 is not a prime
15
[2, 3, 5, 7] 

var primes = [2];
var sum = 0;

// Start looping
for(var i = 3; i <= 10; ++i) {
  var prime = true; // Prime until proven innocent
  for(var j = 0; j < primes.length; ++j) { // Length stays same until later
    if(i % arr[j] === 0) { // The meat of your isPrime function: divisible?
      prime = false;
      break; // Stop the loop early: number is not prime!
    }
  }
  if(prime) { // We have a prime!
    primes.push(i); // Add it to our list of primes
    sum += i; // Add the prime to the sum
  }
}

// Log the result
console.log("The sum of primes up to 10 (inclusive)", sum);
console.log("These primes were", primes);