Why Is This Array Out of Index

Why does OutOfRange array index still prints a value?

What you're seeing is undefined behaviour. C++ doesn't do bounds checking on your array like other languages like Java and C# do. If you give an invalid array index, it still points to a piece of memory, just not memory that you've allocated to the array.

That being the case, if you give an out-of-range array index, it won't throw an exception, it will just do something random (print garbage, segfault, etc). In your case it's not crashing but instead printing the contents of the memory at that index, formatted as an int.

How to fix array index out of bounds error?

This works

for(int i=0;i<fiblist.length;i++){
   System.out.print(fiblist[i]+",");
}
System.out.println();

for (int i=0;i<5;i++){
    temp=fiblist[i];
    fiblist[i]=fiblist[fiblist.length-i-1];
    //the first ellement= the last
    //the second=second from last...
    fiblist[fiblist.length-1-i]=temp;
}

for(int i=0;i<fiblist.length;i++){
    System.out.print(fiblist[i]+",");
}

Output:

1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,
6765,4181,2584,1597,987,8,13,21,34,55,89,144,233,377,610,5,3,2,1,1,

Accessing an array out of bounds gives no error, why?

Welcome to every C/C++ programmer's bestest friend: Undefined Behavior.

There is a lot that is not specified by the language standard, for a variety of reasons. This is one of them.

In general, whenever you encounter undefined behavior, anything might happen. The application may crash, it may freeze, it may eject your CD-ROM drive or make demons come out of your nose. It may format your harddrive or email all your porn to your grandmother.

It may even, if you are really unlucky, appear to work correctly.

The language simply says what should happen if you access the elements within the bounds of an array. It is left undefined what happens if you go out of bounds. It might seem to work today, on your compiler, but it is not legal C or C++, and there is no guarantee that it'll still work the next time you run the program. Or that it hasn't overwritten essential data even now, and you just haven't encountered the problems, that it is going to cause — yet.

As for why there is no bounds checking, there are a couple aspects to the answer:

An array is a leftover from C. C arrays are about as primitive as you can get. Just a sequence of elements with contiguous addresses. There is no bounds checking because it is simply exposing raw memory. Implementing a robust bounds-checking mechanism would have been almost impossible in C.
In C++, bounds-checking is possible on class types. But an array is still the plain old C-compatible one. It is not a class. Further, C++ is also built on another rule which makes bounds-checking non-ideal. The C++ guiding principle is "you don't pay for what you don't use". If your code is correct, you don't need bounds-checking, and you shouldn't be forced to pay for the overhead of runtime bounds-checking.
So C++ offers the std::vector class template, which allows both. operator[] is designed to be efficient. The language standard does not require that it performs bounds checking (although it does not forbid it either). A vector also has the at() member function which is guaranteed to perform bounds-checking. So in C++, you get the best of both worlds if you use a vector. You get array-like performance without bounds-checking, and you get the ability to use bounds-checked access when you want it.

Array Index Out Of Bounds Exception

for(i = 1; i <= height.length; i++){
        heightAsString = JOptionPane.showInputDialog("Please Enter The Height Of Person " + i);
        height[i-1] = Double.parseDouble(heightAsString);

        if(height[i-1] > 1.8){
            over18 += 1;
        }

        if(height[i-1] < 1.6){
            under16 += 1;
        }
    }

use height[i-1], because array index starts from 0.

How do I see if an array index I'm accessing is out of range of the array?

if you have array like this(int[] a = new in[10]) you can test it like

if(index < a.length)

or if you have ArrayList<Integer> b = new ArrayList<Integer>(); you can test it like

if( index < b.size())

java array index out of bounds

You are initializing the array before you initialize the size variable. The size variable has a default value which is passed into the array constructor and sets the array to that size. to fix the problem just move the initialization of the array into the constructor after the size variable is set.

public class RandomArray
{
    /*A class that contains an array of random numbers and methods that output
   the numbers' minimum, maximum and average values. Also includes a method
   that outputs a string representation of the numbers.*/
   int size, min, max;
   String array;
   int[] numbers;

   public RandomArray(int sizeOfArray) {
   Random generator = new Random();

   size = sizeOfArray;

   numbers = new int[size];

   for (int i = 0;i < size;i++)
   {
       numbers[i] = generator.nextInt(size + 1);
   }
}

Also I've noticed a bug with the string output method. The array would be overwritten on each iteration. To solve this you must add the array to itself.

public String toStringArray()//Outputs a string representation of all the numbers in the array
{
   for (int i = 0; i < size;i++)
   {
       array = array + Integer.toString(numbers[i]) + " ";
   }

   return array;
}

for(int : array) index out of bounds exception

You reference arrays by providing an index of the location you want to get the value from. And these indexes start with zero. So to get the first value from the array, you would do bob[0]. What that for-loop is doing is automatically going through each element of the array, and one-by-one, putting the value into j. So to print out the value, you just do System.out.println(j);