Java String.charAt() not working properly

Question

I'm trying to read in a grid from a text file to construct a graph out of it.

Goal nodes are indicated by a X while other nodes are indicated by a ..

I have the number of rows and the number of columns as well. So, I'm basically just trying to read in the next line for each row, then get the character at each column location by passing the iterator for column location and check it against an X to see if it should be marked as a goal node.

goalGraph = new int[rows][cols];
for (int i = 0; i < rows; i++) {
    String readLine = in.nextLine();
    System.out.println(readLine);
    for (int ii = 0; ii < cols; ii++) {
        char c = readLine.charAt(ii);
        if (c == 'x') {
            goalGraph[i][ii] = 1;
        }
        else {
            goalGraph[i][ii] = 0;
            System.out.print(".");
        }
    }
}

But I keep getting an arrayoutofbounds error at index (0).

Another strange thing is that it is not printing readLine when the lower code is implemented, but taking it out so it looks like:

goalGraph = new int[rows][cols];
for (int i = 0; i < rows; i++) {
    String readLine = in.nextLine();
    System.out.println(readLine);
    /**
    for (int ii = 0; ii < cols; ii++) {
        char c = readLine.charAt(ii);
        if (c == 'x') {
            goalGraph[i][ii] = 1;
        }
        else {
            goalGraph[i][ii] = 0;
            System.out.print(".");
        }
    }
    **/
}

As a result, the line being read in are printed, and I get the correct string:

X.....

...X..

......

.X....

Can anyone point me in the right direction?

Thanks!

Can you show us the grid that you are reading from in the text file? — classicjonesynz
– classicjonesynz, Commented Sep 13, 2012 at 1:22
It looks like the "X...." etc... I have in the post without the gaps between the lines. I'm not sure why it's not letting my do multiple lines so I had to hit 'enter' twice. — user1667270
– user1667270, Commented Sep 13, 2012 at 1:27
It just doesn't make any sense to me because if I remove the whole char block as in the second code example, the print statement starts working and shows that the line is indeed being read in correctly. Yet the block being commented out is below the print statement so I have no idea why at least the first print statement doesn't even work when the code isn't commented out. — user1667270
– user1667270, Commented Sep 13, 2012 at 1:30

Keith Randall · Accepted Answer · 2012-09-13 01:46:57Z

1

There is nothing obviously wrong with the code you've shown. We need to see more. Probably what is happening is that there aren't actually rows rows of data in your input. How are rows and cols generated? What type is in, and how is it initialized? What line does the arrayoutofbounds error happen at?

answered Sep 13, 2012 at 1:46

Keith Randall

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4 Comments

user1667270 Over a year ago

I can't seem to put line breaks in for some reason so I apologize for the block. I'm printing 'rows' and 'cols' before entering into this operation and they are correctly being read in so in this instance rows = 5 and cols = 6. They are generated by reading in the first two integer values in the text file. 'in' is the scanner file is initialized as such: Scanner in = new Scanner(new File(fileName)); There isn't actually any line specified, it just says "java.lang.StringIndexOutOfBoundsException: String index out of range: 0"

user1667270 Over a year ago

As above, it just doesn't make any sense because when I comment out all the character operations and such, the line where I print 'readLine', the line being read in from the text file, it does work and shows that it is being read in correctly. Why would the code below the first print operation be affected by what comes after? Should it at least print the read in line once before all the character shenanigans come into play?

Keith Randall Over a year ago

@user1667270: You should really get the full stack trace of your exception - it may not even be generated in the code you think it is. If you run on the command line, it should just print it. If it's a GUI, there may be a log file you can coax out of it that has the full stacktrace.

Keith Randall Over a year ago

@user1667270: If your output is buffered, it is possible that subsequent errors cause previous prints to never make it to the terminal. Try putting System.out.flush() after your prints (this is unlikely to fix it as the default does the flush already - but if your stack traces aren't being printed, something is weird about your I/O configuration).

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