为什么我的FizzBu​​zz代码无法正确输出?

本文介绍了为什么我的FizzBu​​zz代码无法正确输出?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 function counter(numOne, numTwo) { for (let i = 0; i <= 100; i++) { if (i % numOne === 0) { console.log("Fizz"); } if (i % numTwo === 0) { console.log("Buzz"); } if (i % numOne === 0 && i % numTwo === 0) { console.log("FizzBuzz"); } else if (i <= 100 && i !== i % numOne === 0 || i !== i % numTwo === 0) { console.log(i); } } } counter(3, 5);

对于else if循环，它应该console.log所有< = 100 ，但不是 i％numOne === 0 和的数字> i％numTwo === 0 .那么，为什么在输出中仅显示 Fizz ， Buzz 和 FizzBu​​zz ?

For the else if loop, it should console.log all numbers that are <=100, but are not i % numOne === 0 and i % numTwo === 0. So why are only Fizz, Buzz, and FizzBuzz showing up in the output?

推荐答案

好，我不想写一个答案，但是由于您是新来的，所以我将以一种更有意义的方式来说明这一点:

Ok, I didn't want to write an answer but since you're new here, I'll put this in a more meaningful way:

function counter(numOne, numTwo) { for (let i = 0; i <= 100; i++) { const isFizz = i % numOne === 0 const isBuzz = i % numTwo === 0 if (isFizz && isBuzz) { console.log("FizzBuzz"); } else if (isFizz) { console.log("Fizz"); } else if (isBuzz) { console.log("isBuzz") } else { console.log(i); } } } counter(3, 5);

在您的示例中，您有:

i !== i % numOne === 0

如上所述，这里有两个问题:

as stated above, there are two issues here:

i！== i 永远不能为 true ，它的值相同，始终为 i === i 或您的情况错误

由于上述错误，因此您将具有以下数学公式: false％numOne 这将导致 NaN ，并且 NaN不等于0

i !== i can never be true, it's the same value, it's always i === i or in your case false

Since the above is false, you'll have a math equation of: false % numOne this will result in a NaN and NaN does not equal 0

希望如此，以上评论有助于理解您的问题

Hope this and the comments above helps understand your issue