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Danny Short
Courses Plus Student 2,470 PointsWhy does this always evaluate as false?
My Solution doesn't work with my if statement i used typeof to compare the type of my variable against the type number
// Collect input from a user
let userNumber = prompt('Choose a number?');
// Convert the input to a number
const highNumber = parseInt(userNumber);
if (typeof highNumber === Number) {
// Use Math.random() and the user's number to generate a random number
const randomNumber = (Math.random() * (highNumber) +1 );
// Create a message displaying the random number
console.log(`${parseInt(randomNumber)} is a random Number between 1 and ${highNumber}`);
} else {
console.log(`That's not a number you trickster`);
}
2 Answers
Steven Parker
231,248 PointsThe "typeof" operator will return a string with the type in lower case:
if (typeof highNumber === "number") {
Also, "randomNumber" is already a number, you don't need to perform "parseInt" on it.
Danny Short
Courses Plus Student 2,470 PointsI think I did the parse Int because random number is a float at that point & wanted it to be an int could use Math.floor in the original declaration instead
Steven Parker
231,248 PointsSince "parseInt" is specifically for strings, an implicit coercion has to happen here. The use of "floor" is a much more conventional approach.