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- 09-21-2009, 03:32 PM #1
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Input technique for unknown lines of input
I have a problem with input techniques from user. Suppose user can feed up program 3 lines as:
Java Code:Department of Computer science and engineering University of Dhaka
Java Code:Department of Computer science and engineering University of Dhaka
Java Code:import java.util.ArrayList; import java.util.Iterator; import java.util.Scanner; public class InputTechnique { public static void main(String args[]){ Scanner scan = new Scanner(System.in); ArrayList<String> aList = new ArrayList<String>(); while(scan.hasNextLine()){ aList.add(scan.nextLine()); } Iterator it = aList.iterator(); while(it.hasNext()) System.out.println(it.next()); } }
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How do you know when the user is done inputting text? Do they send a blank line?
- 09-21-2009, 03:39 PM #3
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What do you mean by "it hangs"?
It'll accept your input, so it's not hanging...you just haven't given it an exit.
- 09-21-2009, 03:52 PM #4
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i mean it cannot break the first while loop and continue taking inputs from user.
User can feed blank lines as
Java Code:hello i am new to this forum
Java Code:hello i am new to this forum
- 09-21-2009, 03:56 PM #5
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So, as Fubarable has asked, and (I think) was pointed out in the thread you tried to hijack, how does the user exit?
- 09-21-2009, 04:01 PM #6
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yes, my question is how does the program terminate? Is there any process to solve this. I need this requirement for solving a UVA online judge problem, and if i am failed to make you understand what i am trying to say, just visit uva online judge and the problem ID is 10815[Andy's First Dictionary].
Thanks
- 09-21-2009, 04:09 PM #7
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OK...step back then...how do you want the user to say they've entered enough?
In the real world I have never written a command line user input program...at least not in the past 15+ years. So, how do you want them to inform the program they have finished?
Fubarable has given an option in his post, and there were one or two on the other thread.
- 09-21-2009, 04:33 PM #8
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How can i find the End Of Files? Would you like to mention the technique.
- 09-21-2009, 04:42 PM #9
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What end of file?
It's user input.
You have to decide what the user needs to enter to say they've finished.
There is no "technique"...what does the user have to do to say they've finished entering data?
- 09-21-2009, 04:48 PM #10
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Ok, Thanks.
- 09-21-2009, 04:52 PM #11
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Right, no problem.
Do you now understand what you have to do to get this to work?
- 09-22-2009, 04:12 AM #12
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Yes, thanks a lot Tolls. I have to take the input from files rather than console.
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At least for me, your creeping requirements have confused me greatly and continue to confuse me. In your first post, you mention nothing about files and fact state this:
I have a problem with input techniques from user. Suppose user can feed up program 3 lines as:
Java Code:Scanner scan = new Scanner(System.in);
So I have to ask, what the heck is going on? What are your real requirements here and why are you changing things?
If you confuse your questions you will get nothing but confusing answers.Last edited by Fubarable; 09-22-2009 at 04:42 AM.
- 09-22-2009, 09:13 AM #14
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- 09-22-2009, 10:34 AM #15
Hi,
U just take the input
"y" or "n" from the user .If they say "n" just come out.
-Regards
RamyaRamya:cool:
- 09-22-2009, 06:20 PM #16
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Okay, i am explaining now. The first thing is that i am new in java. Actually i like C++. Look at this code bellow
Java Code:#include<iostream> #include<cstring> using namespace std; struct x{ char array[100]; }sample[100]; int main(){ freopen("input.txt", "r", stdin); char input[101]; int indexx = -1; while(gets(input)){ strcpy(sample[++indexx].array, input); } for(int i = 0; i <= indexx; i++) cout << sample[i].array; return 0; }
Java Code://freopen("input.txt", "r", stdin);
i wanted to write this code in java as
Java Code:import java.util.ArrayList; import java.util.Iterator; import java.util.Scanner; public class InputTechnique { public static void main(String args[]){ Scanner scan = new Scanner(System.in); ArrayList<String> aList = new ArrayList<String>(); while(scan.hasNextLine()){ aList.add(scan.nextLine()); } Iterator it = aList.iterator(); while(it.hasNext()) System.out.println(it.next()); } }
Now i think, you understand my words what i am trying to say. Now give me a solution.
- 09-23-2009, 09:26 AM #17
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And this is the problem.
Java is not C++.
You cannot simply take code from one language and expect to turn it into another one without actually learning the language itself.
I am, however, not going to give you the solution. There is no "freopen" function in Java.
Scanner is not freopen.
You need to learn how Scanner works...especially with System.in.
As has been said several times on this thread, you need to decide what value the user has to enter (and it will involve <enter> since that's how nextLine works) to indicate they have finished.
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