C# – How to Generate Random Number Between 0 and 1

.netc++random

I want to get the random number between 1 and 0. However, I'm getting 0 every single time. Can someone explain me the reason why I and getting 0 all the time?
This is the code I have tried.

Random random = new Random();
int test = random.Next(0, 1);
Console.WriteLine(test);
Console.ReadKey();

Best Answer

According to the documentation, Next returns an integer random number between the (inclusive) minimum and the (exclusive) maximum:

Return Value

A 32-bit signed integer greater than or equal to minValue and less than maxValue; that is, the range of return values includes minValue but not maxValue. If minValue equals maxValue, minValue is returned.

The only integer number which fulfills

0 <= x < 1

is 0, hence you always get the value 0. In other words, 0 is the only integer that is within the half-closed interval [0, 1).

So, if you are actually interested in the integer values 0 or 1, then use 2 as upper bound:

var n = random.Next(0, 2);

If instead you want to get a decimal between 0 and 1, try:

var n = random.NextDouble();

Related Solutions

C# Random – How to Generate a Random Number in C#

The only thing I'd add to Eric's response is an explanation; I feel that knowledge of why code works is better than knowing what code works.

The explanation is this: let's say you want a number between 2.5 and 4.5. The range is 2.0 (4.5 - 2.5). NextDouble only returns a number between 0 and 1.0, but if you multiply this by the range you will get a number between 0 and range.

So, this would give us random doubles between 0.0 and 2.0:

rng.NextDouble() * 2.0

But, we want them between 2.5 and 4.5! How do we do this? Add the smallest number, 2.5:

2.5 + rng.NextDouble() * 2.0

Now, we get a number between 0.0 and 2.0; if you add 2.5 to each of these values we see that the range is now between 2.5 and 4.5.

At first I thought that it mattered if b > a or a > b, but if you work it out both ways you'll find it works out identically so long as you don't mess up the order of the variables used. I like to express it with longer variable names so I don't get mixed up:

double NextDouble(Random rng, double min, double max)
{
    return min + (rng.NextDouble() * (max - min));
}

Random Number Generation in C# – Methods and Examples

The default seed is taken from the system clock.

I'm guessing that your GetRandomRectangle method was being called in quick succession and instantating a new instance of Random each time. When you do this, each instance of Random will take the same seed from the system clock, which is why your method created the same rectangle each time.

One solution is to create one instance of Random and pass that into your method:

Random rng = new Random();

Rectangle foo = GetRandomRectangle(rng);
Rectangle bar = GetRandomRectangle(rng);
Rectangle baz = GetRandomRectangle(rng);

// ...

public Rectangle GetRandomRectangle(Random rng)
{
    // create the rectangle using rng
}

Best Answer

Related Solutions

C# Random – How to Generate a Random Number in C#

Random Number Generation in C# – Methods and Examples

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