我正在尝试解决的初学者 Javascript 函数问题

Question

The question I'm trying to solve goes as follows:

I often struggle to know whether I should wear shorts or pants on a given day. Please help me decide by writing me a function called isShortsWeather

It should accept a single number argument, which we will call temperature (but you can name it whatever you want).

• If temperature is greater than or equal to 75, return true .
• Otherwise, return false
• This exercise assumes temperature is in Fahrenheit

Expected result:

isShortsWeather(80) //true
isShortsWeather(48) //false
isShortsWeather(75) //true

The code I wrote is:

function isShortsWeather(temperature) {
    if (temperature < 75); {
        return false;
    } 
    if (temperature >= 75) {
        return true;
    }
}

As snippet:

 function isShortsWeather(temperature) { if (temperature < 75); { return false; } if (temperature >= 75) { return true; } } console.log(isShortsWeather(80)) //true console.log(isShortsWeather(48)) //false console.log(isShortsWeather(75)) //true

Please help me out by informing me what is wrong with my code and how should I fix this issue. I feel like I am relatively close to solving it. Thanks!

Answer 1

只需返回布尔值：

return temperature >= 75;

Answer 2

the main reason it's not working is because you have a extra ; after the first condition. you can shorten the function body to one line

function isShortsWeather(temperature) {
    return temperature >= 75;
}

Answer 3

you forgot ; at the line 2, if you remove it it will work. Also it would be better if you make the 2nd if statement in else if

    function isShortsWeather(temperature) {
        if (temperature < 75) {
            return false;
        } else if (temperature >= 75) {
            return true;
    }

Answer 4

I would recommend a return on both if statements and not a console log on false. You will use console.log to call the function. I have edited the code a bit as the semicolon on line 2 not needed.

 function isShortsWeather(temperature) { if (temperature < 75) { return false; } else { return true; } } temperature = 74; console.log(isShortsWeather(temperature));

Answer 5

Below you will find the correct code that is working. Hope answer your questions.

 function isShortsWeather(temperature) { if (temperature >= 75) { return true; } else { return false; } } console.log(isShortsWeather(76)); console.log(isShortsWeather(74));

Answer 6

You can return the value instead of console logging it.

It would also be a good idea to refactor the "magic number" 75 into a variable with a default value, allowing you to easily change or override it later.

function isShortsWeather(temperature, cutoffTemp = 75) {
  if (temperature < cutoffTemp) {
    return false;
  } 
  if (temperature >= cutoffTemp) {
    return true;
  }
}

console.log(isShortsWeather(81));

Answer 7

function isShortsWeather(temperature) {

if (temperature < 75); {return 0;}

if (temperature >= 75) {return 1;}
}

I think that will solve your problem