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Start your free trialRobert Hubbard
Full Stack JavaScript Techdegree Student 6,780 PointsStrangely a variable is not defined on the same line, even though it was used earlier on the line.
When defining randomNum, error "Uncaught ReferenceError: userLowInt is not defined" occurs. userLowInt is recognized the first time, but not the 2nd.
// Collect input from a user
var userLowInput = prompt("Give me a low number");
var userHighInput = prompt("Give me a high number");
//console.log(`1) ${userLowInput}`);
if(userLowInput && userHighInput){
// Convert the input to a number
const userlowInt = parseInt(userLowInput);
const userHighInt = parseInt(userHighInput);
//console.log(`2) ${userInt}`);
// Use Math.random() and the user's number to generate a random number
var randomNum = Math.floor(Math.random() * (userHighInt - userlowInt+1) ) + userLowInt;
console.log(`3) ${randomNum}`);
// Create a message displaying the random number
console.log(`${randomNum} is the random number between ${userLowInt} and ${userHighInt}`);
}else {
console.log(`You need to add a number`);
}
1 Answer
Steven Parker
231,261 PointsThe variable is declared and used the first time as "userlowInt" (with a lower case "l"), but the undefined reference is for "userLowInt" (with a capital "L").