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Start your free trialStacy Loggins
3,632 PointsparseInt()? Why doesn't the parseInt() work inside the function?
function randomNumber(num1, num2) { //Why doesn't the parseInt() work inside the function? var number =Math.floor(Math.random() * (parseInt(num1) - parseInt(num2) + 1)) + num2; return number; }
// parseInt() outside the function works even with the parseInt() inside the function
/var upper = parseInt(prompt("what is the larger number")); var lower = parseInt(prompt("what is the smaller number")); console.log(randomNumber(upper, lower));/
//This is the code that will not parse the information in the function
/var upper = prompt("what is the larger number"); var lower = prompt("what is the smaller number"); console.log(randomNumber(upper, lower));/
2 Answers
Steven Parker
231,271 PointsI tried both samples and they each seem to work. I did notice that the inside-the-function version relies on implicit type coercion for the last part of the formula ("+ num2
") where it might be better practice to apply parseInt again to convert it explicitly. But the implicit conversion does work.
If you're still having trouble, it might be helpful to share this as a workspace "snapshot", which would make it possible to exactly replicate the entire environment.
Obe Juarez
6,357 PointsYou don't need parseInt because your not getting a number from the user. if you did var userNumber= prompt('type a number'); THEN you would need to parseInt the number so it sees it as a number and not a string . But since you arent asking for a number ... you dont need it . The computer is generating one for you.