Exiting Loop After a Delay Help!

[heroes100]

What is it you are trying to achieve?
I am trying to get a loop to stop after a delay of w.e time.

What is the problem you are having?
I have a loop which generates random symbols and i would like it to stop generating these symbols after a designated time limit.

Describe what you have tried to solve this problem
I have tried to create a variable and make it add to the variable every loop and once it ='s a number it stopes but this will not work for i get a syntax error. :/

Post any relevant code (You may choose to attach the file instead of posting the code if it is too long)

Turing:

loop let := Rand.Int (1,122) put chr (let), "".. end loop end loading

Please specify what version of Turing you are using
4.11 (or i can also load it with 4.0.5)

[BigBear]

You have no exit condition

I think you might want to use a for loop

[heroes100]

How can i make "let" a random int between 1-122 in a for loop?
Also when it is a for loop how can i exit it with a delay?? :/

[Kharybdis]

Turing:

var let : int var count := 1 loop     randint (let, 1, 5) % (integer value, lowest, highest)     count += 1     put let     exit when let = 2 % exit condition if let = 2 end loop put "let appeared, ", count, " times."

[heroes100]

Thanks for the responses. I have found a way around this. Even though the main question of how you can make a loop stop after a delay hasn't been answered but i have solved the problem. So all is well thx.

[Superskull85]

What do you mean by "stop after a delay?" Do you mean something like this:

Turing:

loop    let := Rand.Int (1,122)    put chr (let), ""..    delay (100)    exit end loop

Or do you mean like what Kharybdis posted:

Turing:

var let : int var count := 1 loop     randint (let, 1, 5) % (integer value, lowest, highest)     count += 1     put let     exit when let = 2 % exit condition if let = 2 end loop put "let appeared, ", count, " times."

Or do you means something like this:

Turing:

var let : int var count : int := 0 loop     let := Rand.Int (1, 122)     count += 1     put chr (let), "" ..     exit when count = 10 end loop

I am a little confused about what you mean.

Anyways, if you wanted to use a for loop with a random upper bound than you could do something like:

Turing:

for i : 1 .. Rand.Int (1, 222)     put chr (i) .. end for

What did you do to solve the problem?

[heroes100]

I meant more of the first but i do not know if that will still generate enough chars. as i need. I have ended up doing the 3rd thing you posted though but i was wanting it to generate random chars. and then stop after a set delay. But there is no need atm because i fixed it.

[B-Man 31]

if your trying to do it after a set amount of time, you could do something like this, more or less:

Turing:

var timepassed : int loop     timepassed := Time.Elapsed % starts a timer function     let := Rand.Int (1, 122)     put chr (let), "" ..     if Time.Elapsed - timepassed >= 3000 then % the 3000 represents 3 seconds, change it to whatever you want         exit %so basicly, if more than 3 seconds passed, then exit the loop     end if end loop