Welcome to the Treehouse Community
Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.
Looking to learn something new?
Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.
Start your free trialAnouk Lejeune
3,505 PointsWhat is wrong with next code?
function max (number1, number2) { if (parseInt(number1) > parseInt(number2)){ var one= number1; return one; } else { var two = (parseInt(number2) > parseInt(number1)); return two; }
function max (number1, number2) {
if (parseInt(number1) > parseInt(number2)){
var one= number1;
return one;
} else {
var two = (parseInt(number2) > parseInt(number1));
return two;
}
3 Answers
Ayoub AIT IKEN
12,314 Pointshey ! you dont need the parseInt function, you didnt close your brackets of the 'else', and you can write your method more easily like that :
function max(number1, number2)
{
if (number1 > number2)
{
return number1;
}else
{
return number2;
}
}
Anouk Lejeune
3,505 PointsHey Ajoub,
Thanks for the answer and reminding me of the missing closing bracket. But I still have one question. In Your code, what happens with the case where number1 = number2? Because know it looks like it will just return number2, when they're equal. And it was only allowed to give the greatest number, not?
Ayoub AIT IKEN
12,314 Pointshey ! my answer only depend on the challenge question, it is all what is asked, if you want to manage the equal case , you have to add it in the if else to:
function max(number1, number2)
{
if ( number1 === number2)
{
alert("numbers are equal");
}
else if (number1 > number2)
{
return number1;
}else
{
return number2;
}
}