int array[] = {0,0,0,0,0,0,0,0,0,0}
if (array[9] == 0) {  //if the array is full, skip this
     currentVal = array[i]; //store current value in array
     i++;
} else {
     //store max value of array in medHigh
     for (int x = 0; x < 10; x++) {   
        if (array[x] > medHigh) {
        medHigh = array[x];
        }
     } 
}
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        Oliver Klingefjord
        
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                    2What's `i`? What's its value? – Biffen Oct 12 '16 at 10:15
1 Answers
1
            Simple because your array has length 10 so the maximum index is 9. Arrays in Java are 0-based. It means, that it's indexes are from 0 incl to array.length - 1 incl which in your case is 9. 
Java gave you a stacktrace. It looks like following (by the way, A L W A Y S attach it to the question about Exceptions):
java.lang.ArrayIndexOutOfBoundsException: 10
at package.ClassName.methodName(ClassName.java:lineNumber)
...
Is the line number the number of line in which you call currentVal = array[i];? If yes, it's your answer - i is 10.
If you paste your stack trace, comment my answer for further help. I'll answer with pleasure.
 
    
    
        xenteros
        
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