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JavaScript JavaScript Loops, Arrays and Objects Simplify Repetitive Tasks with Loops A Closer Look at Loop Conditions

Faraz Khalid
Faraz Khalid
4,077 Points
Posted by Faraz Khalid
Faraz Khalid
4,077 Points

Why can't the random number for variable guess be declared outside the function?

Why can't the random number for the variable guess be generated outside the function as follows >

var ranNum= randomNumber (1000); var guess =randomNumber (1000); var counter =1;

while (guess !== ranNum) { counter +=1; }

immad Ud din
.a{fill-rule:evenodd;}techdegree
immad Ud din
Front End Web Development Techdegree Student 14,756 Points

If you initialize guess outside it will have only one value as var guess = randomNumber(1000) will run only once. say guess has 456 and randomNumber has 500. so the two values will not match thus makes the loop endless. for the sake of guess to get a fresh value it is inside the loop until it matches the randomNumber.

1 Answer

Steven Parker
Steven Parker
229,657 Points
Steven Parker
Steven Parker
229,657 Points

If both numbers are generated before the loop, then unless they happen to be the same (which is extremely unlikely!) the loop would never end.

The purpose of the exercise is to generate a new number inside the loop to see how many tries it takes to match the one that was generated before the loop started.