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JavaScript JavaScript Basics (Retired) Working With Numbers Create a random number

Posted by furkan
furkan
11,733 Points

Can't see any error, but random number generator won't work?

I used the following code:

var dieRoll = Math.floor(Math.random() * 6) + 1; dieRoll;

The final result always returns 3 as the "random number". The random generator does not work. Any advice?

2 Answers

Alan Jaffery
Alan Jaffery
4,803 Points
Alan Jaffery
Alan Jaffery
4,803 Points

Just checked my self. With this code it works: var dieRoll = Math.floor(Math.random() * 6) + 1; alert(dieRoll);

Try it your self. After refreshing the page i got always another (random) number

furkan
furkan
11,733 Points

Even with this code, if I run the first part, create a random number and store that to the variable. Then press enter, it provides an undefined variable. Then run the next line of code alert=(dieRoll), and still gives a single value and doesn't change each time the variable is called. Your code returns 6 each time.

furkan
furkan
11,733 Points

Thank-you for the clarification

Alan Jaffery
Alan Jaffery
4,803 Points
Alan Jaffery
Alan Jaffery
4,803 Points

No problem. Hope i could help you :)

furkan
furkan
11,733 Points
furkan
furkan
11,733 Points

That is the full code. It was a test run to see how random numbers work. I created the variable dieRoll then added the Math.floor and Math.random functions, and finally called the variable dieRoll to provide the answer, like this:

var dieRoll = Math.floor(Math.random() * 6) + 1; dieRoll;

Strange enough, I tried running the above code as it is on console.log in chrome, and it works perfectly fine, however, if I try to run the random number by storing the number in the variable dieRoll pressing enter (which gives me undefined as a result), and then try to run the dieRoll; as a separate command on the next line, it gives me the same number each time, 3 as the result.

Alan Jaffery
Alan Jaffery
4,803 Points
Alan Jaffery
Alan Jaffery
4,803 Points

Thats because the variable is !once! filled and after that not changed anymore. Due to that each to giving out the content of the variable will be the same over and over. you would need to set it each time new to get other numbers. try the code i answered. With that the random number will always apear newly

kind regards & sorry for my bad english :) alan