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JavaScript Introducing the Practice

Samantha Atkinson
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Samantha Atkinson
Front End Web Development Techdegree Student 36,955 Points
Posted by Samantha Atkinson
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Samantha Atkinson
Front End Web Development Techdegree Student 36,955 Points

Why do I have to convert the user's number to a floating-point number?

Am I overthinking this? It seems to me I'm supposed to create some maths with 2 variables with integer numbers right? So why does it say in the comments:

//4. Convert that value from a string to a floating-point number

Shouldn't that be an integer number? Is it because if the user enters an integer number parseFloat returns an integer number anyway and if they do input a decimal number pareseFloat will do its thing?

1 Answer

Steven Parker
Steven Parker
229,732 Points
Steven Parker
Steven Parker
229,732 Points

Since the numbers will be used to do math, the program should accept anything entered.

And it sounds like you nearly figured it out yourself, as you pointed out, a "parseFloat" will work with any number. But a "parseInt" would not correctly handle a non-integer, so it makes sense to convert the input as a floating-point number.

Samantha Atkinson
seal-mask
.a{fill-rule:evenodd;}techdegree
Samantha Atkinson
Front End Web Development Techdegree Student 36,955 Points
Samantha Atkinson
.a{fill-rule:evenodd;}techdegree
Samantha Atkinson
Front End Web Development Techdegree Student 36,955 Points

Hi Steven, thank you for your reply. You've been kind enough to reply to previous questions of mine, thank you!

I'm sorry your last confused me.

"And "parseFloat" always returns a floating-point number, even if the input has no explicit decimal point."

I think why I'm confused is because I did a console.log and expected to see my integer number as 0.5 when I had input 5. So are you saying even an integer number is still considered a floating-point number by parseFloat regardless?

Steven Parker
Steven Parker
229,732 Points
Steven Parker
Steven Parker
229,732 Points

I confused myself for a moment! :see_no_evil: But it's the conversion process that's important:

console.log(parseFloat("5.2"));   // outputs: 5.2
console.log(parseInt("5.2"));     // outputs: 5