Build a Leap Year Calculator - Build a Leap Year Calculator

JavaScript
1

Tell us what’s happening:

Hi, I can’t pass step 10 but result seems ok, could you help to check for me? Thanks.

Your code so far

const year = 0;
function isLeapYear(year){

if(year %4 === 0 && year %400 === 0){
return year + " is a leap year."} 

else if (year %4 === 0 && year %100 === 0){
  return year + " is not a leap year."
  } 
  
else if (year %4 === 0){
  return year + " is a leap year."
} 

else{
  return year + " is not a leap year.";
}


};
const result = isLeapYear(2000);

console.log(result);

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/143.0.0.0 Safari/537.36 Edg/143.0.0.0

Challenge Information:

Build a Leap Year Calculator - Build a Leap Year Calculator

2

Hi @timduncan0322

8. You should call the isLeapYear function and pass year as a parameter.

Are you sure you passed test 8?

Happy coding

3

yes, just miss step 10

4

Where you do define the year parameter? Which line of code?

5

I declare “year” in the 1st line (const year = 0;), and define it in the if statements, I not not sure if it is correct.

6

On which lines do you access that variable?

7

i think through the return lines, I also tried with the following, but still the same.

if(year %4 === 0 && year %400 === 0){

year = year + " is a leap year."

return year;

} else if (year %4 === 0 && year %100 === 0){

year = year + " is not a leap year."

return year;

} else if (year %4 === 0){

year = year + " is a leap year."

return year;

} else{

year = year + " is not a leap year.";

return year;

}

8

if Doesn’t define a variable.

You define year twice:

const year = 0;
function isLeapYear(year){

The first line, as you mentioned, and as a function parameter.

Can you think of how this affects scope?

9

got it, shouldn’t declare in the beginning, should be for testing only, thanks.

10

This is sufficient for testing. You should pretty much never define a variable that’s used as a parameter outside of the function.

11

yeah, but the tests want a year variable defined in the global scope

1 Like
12

In that case the function should be called with the variable and the parameter should have a different name, right?

13

you can open an issue on github for this if you want

but a global variable with a name, and a function parameter with the same name, while confusing, don’t interact with each other

1 Like
14

Wow, ok I really misunderstood that, thanks. It was really just the argument that was the problem.