Array.BinarySearch does not find item using IComparable

Question

If a binary search requires an array to be sorted before hand, why does the following code work?

string[] strings = new[] { "z", "a", "y", "e", "v", "u" };
int pos = Array.BinarySearch(strings, "Y", StringComparer.OrdinalIgnoreCase);           
Console.WriteLine(pos);

And why does this code result return -1?

 public class Person : IComparable<Person> {

    public string Name { get; set; }
    public int Age { get; set; }


    public int CompareTo(Person other) {
        return this.Age.CompareTo(other.Age) + this.Name.CompareTo(other.Name);
    }
}
var people = new[] {

                new Person { Age=5,Name="Tom"},
                new Person { Age=1,Name="Tom"},
                new Person { Age=2,Name="Tom"},
                new Person { Age=1,Name="John"},
                new Person { Age=1,Name="Bob"},
            };


            var s = new Person { Age = 1, Name = "Tom" };

            // returns -1
            Console.WriteLine(
                Array.BinarySearch(people, s)
            );

The first part most likely works because there's an even number of elements in the array. Binary search tries checks the element in the "middle." In this case there are two elements in the middle so it probably picks the leftmost one, which happens to be Y, and returns the correct position (2). Put an another letter (other than Z) after Y and you'll get a negative result. — Roman
– Roman, Commented May 29, 2010 at 4:26

Matthew Flaschen · Accepted Answer · 2010-05-29 04:32:32Z

2

If you violate the pre-condition ("The elements of array must already be sorted in increasing value according to the sort order defined by comparer; otherwise, the result might be incorrect."), you cause undefined behavior, which sometimes includes, "Hey, it looks like it worked." The same is true with use after free and many other programming errors.

I'm not surprised it works for that input, because "y", and "e" are both equally middle elements. It uses the lower middle, "y", which matches. If you try:

string[] strings = new[] { "z", "a", "y", "e", "e", "v", "u" };

you'll see that doesn't work. Note that there are an odd number of elements. The target is before the middle, while it should be after.

edited May 29, 2010 at 4:32

answered May 29, 2010 at 4:23

Matthew Flaschen

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