这个条件在Prime函数中的输出是什么？ [英] What is the output of this condition in the Prime Function?

问题描述

我从另一篇文章中发现了这段代码，我试图理解这个解决方案的一部分。

 
（$） 
 函数isPrime（num）{
 for（ var i = 2; i< num; i ++）{
 if（num％i === 0）{
 return false; 
} 
} 
返回true; 
} 
  
尤其是这条线
  num％i === 0 
  
什么我可以理解的是， arr 是一个数组，它在2之后产生每个奇数（例如，[2,3,5,7,9,11,13,15， 17]）。然后每个数字通过函数isPrime 运行。我试图理解的是 num％i === 0 与有关的输出是什么（var i = 2; i < num; i ++）？
 
 
 
是这样的输出吗？ 
 li 3（num）％2（i）
 li 5（num）％3（i） 
 
 7（num ）％4（i）
 li 9（num）％5（i）
 li 11（num）％6（i）
 
 
 
 
解决方案
 
 
 
  console.log（`$ {num}％$ {i} === $ {num％i}`）; //额外行
 if（num％i === 0）{
 return false; 
 
 
 
 
 $ b $ p 
 $ b 
在这段代码中，代码实际上做的是它会经过 2 和 num 之间的每一个数字，将它赋值给变量 i ，并检查 num 是否可被 i 整除。如果是，那么它将返回 false 。
 
 
 
 ％函数（称为模函数）基本上取两个数字，并返回除以第二个数字后的第一个数字的余数。举例来说：
  5％2 // = 1，5/2 = 2 with 1超过
 7％3 // = 1，7/3 = 2还剩1美元b $ b  
如果余数是 0 ，那么显然第一个数字可以被第二个数字整除，因为没有什么可以留下。因此  num％i === 0 正在检查可分性，本质上它检查 num 可被 i整除。 
 
 
例如，当检查 5 （带有额外的 console.log 行），这是输出到控制台的内容：
 
 
 < pre class =lang-js prettyprint-override>  5％2 === 1 
 5％3 === 2 
 5％4 === 1 
  
这就是 6 输出为 num ：
  6％2 === 0 
  
（已停止，因为 false 被返回。）
 
I found this code from another post and I'm trying to understand a part of this solution.



function sumPrimes(n) {
  function isPrime(num) {
    for ( var i = 2; i < num; i++ ) {
        if ( num % i === 0 ) {
            return false;
        }
    }
    return true;
}
    var arr = 2;
    for ( var i = 3; i <= n; i+=2 ) {
        if ( isPrime(i) ) {
            arr += i;
        }
    }
return arr;
}
console.log(sumPrimes(10));





The part I'm asking about is this particular function
function isPrime(num) {
        for ( var i = 2; i < num; i++ ) {
            if ( num % i === 0 ) {
                return false;
            }
        }
        return true;
    }
especially this line in question
num % i === 0
What I can understand is that arr is an array that product every odd number after 2 (for example, [2, 3, 5, 7, 9, 11, 13, 15, 17]). Then every number is run through function isPrime. What I'm trying to understand is what are the outputs of num % i === 0 in relation to for ( var i = 2; i < num; i++ )?

Are the outputs like this?


3(num) % 2(i)
5(num) % 3(i) 
7(num) % 4(i) 
9(num) % 5(i) 
11(num)% 6(i)

解决方案


for ( var i = 2; i < num; i++ ) {
    console.log(`${num} % ${i} === ${num % i}`); // Extra line
    if ( num % i === 0 ) {
        return false;
    }
}
In this code, what the code is actually doing is it's going through every single number between 2 and num, assigning it to the variable i, and checking if num is divisible by i. If it is, then it'll return false.

The % function (called the modulo function), basically takes two numbers, and returns the first number's remainder when divided by the second number. So, for example:
5 % 2 // = 1, 5/2 = 2 with 1 left over
7 % 3 // = 1, 7/3 = 2 with 1 left over
If the remainder is 0, then obviously the first number is divisible by the second number, since there's nothing left over. So the line num % i === 0 is checking for divisibility, essentially - it's checking if num is divisible by i.

For example, when checking 5 (with the extra console.log line), this is what's outputted to the console:
5 % 2 === 1
5 % 3 === 2
5 % 4 === 1
And this is what's outputted with 6 as num:
6 % 2 === 0
(It's stopped, because false is returned.)

                        
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