Let's look at a cleaner version of the loop you constructed:
for (i = 0; i < a.length; i++); {
if (a[i] < a[i + 1]) {
return true;
}
else {
return false;
}
}
I should first point out the syntax error in the original loop. Namely, there is a semicolon (;
) before the curly brace ({
) that starts the body of the loop. That semicolon should be removed.
Also note that I reformatted the white-space of the code to make it more readable.
Now let's discuss what happens inside your loop. The loop iterator i
starts at 0
and ends at a.length - 1
. Since i
functions as an index of your array, it makes sense pointing out that a[0]
is the first element and a[a.length - 1]
the last element of your array. However, in the body of your loop you have written an index of i + 1
as well. This means that if i
is equal to a.length - 1
, your index is equal to a.length
which is outside of the bounds of the array.
The function isSorted
also has considerable problems as it returns true the first time a[i] < a[i+1]
and false the first time it isn't; ergo it does not actually check if the array is sorted at all! Rather, it only checks if the first two entries are sorted.
A function with similar logic but which checks if the array really is sorted is
public static boolean isSorted(int[] a) {
// Our strategy will be to compare every element to its successor.
// The array is considered unsorted
// if a successor has a greater value than its predecessor.
// If we reach the end of the loop without finding that the array is unsorted,
// then it must be sorted instead.
// Note that we are always comparing an element to its successor.
// Because of this, we can end the loop after comparing
// the second-last element to the last one.
// This means the loop iterator will end as an index of the second-last
// element of the array instead of the last one.
for (int i = 0; i < a.length - 1; i++) {
if (a[i] > a[i + 1]) {
return false; // It is proven that the array is not sorted.
}
}
return true; // If this part has been reached, the array must be sorted.
}
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