Incorrect Arguments for .Next(...)

This is a series on common ways that System.Random is misused, based on my experience reviewing code test submissions from job candidates.

The most basic error in using System.Random is to get the parameters for the Next method wrong. Part of our coding tests involves generating a random key value from key0 to key9.

Twenty-two of the submissions reviewed used the following incorrect method call to get that random digit:

rnd.Next(0, 9);

What’s the Problem?

You may have heard that there are 2 hard problems in Computer Science: naming things, cache invalidation, and off-by-one errors. This is one or the latter.

The above call will return a number from 0 to 8, inclusive. That 9 in the method call is the exclusive upper bound of the range for random number generator. This is confusing if you just look at the parameter names (maxValue in this case), but is clear in Microsoft’s documentation.

The Random.Next method has three overloads:

int Next()
int Next(int maxValue)
int Next(int minValue, int maxValue)

The first two are equivalent to Next(0, Int32.MaxValue), and Next(0, maxValue), respectively. The docs state that minValue is the “inclusive lower bound” and maxValue is the “exclusive upper bound”, so the numbers returned will satisfy the relation: minValue <= value < maxValue.

The Solution

When using .Next with range arguments you need to remember that the maxValue parameter is the exclusive upper bound, so it needs to be one more than the actual maximum value to want to produce.

So for generating our keys in the range of 0-9 inclusive we need either of:

rnd.Next(10)
rnd.Next(0, 10)

To simulate a six-sided die you would do:

rnd.Next(1, 7);

Next time I’ll talk about pitfalls in seeding the random number generator.